Contoh Soal 4 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 16.& \mathbf{\text{(UN 2015 Matematika IPA)}}\\ & \text{Hasil}{\displaystyle \int 6x{(1-x^2)}^{4}dx\text{adalah ....}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{5}{(1+x^2)}^5+C}\\ \\ \text{b}.{\displaystyle \frac{2}{5}{(1+x^2)}^5+C}\\ \\ \text{c}.{\displaystyle -\frac{1}{5}{(1-x^2)}^5+C}\\ \\ \text{d}.-{\displaystyle \frac{2}{5}{(1-x^2)}^5+C}\\ \\ \text{e}.-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Alternatif 1}\\ & \text{Dengan integral substitusi}\\ & \begin{array}{c}\\ \begin{array}{rl}{\displaystyle }& \int 6x{(1-x^2)}^{4}dx\\ & =\int {(1-x^2)}^4.6xdx\\ & \{\begin{array}{ll}u& =1-x^2\\ & \\ du& =-2xdx\Rightarrow -3dx=6xdx\end{array}\\ & {\displaystyle \int u^4.(-3du)}\\ & =-3{\displaystyle \int u^{4}du}\\ & =-{\displaystyle \frac{3}{5}{u}^5+C}\\ & =-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& {\displaystyle \int {(1-x^2)}^{4}6xdx}\\ & =\int {(1-x^2)}^2.(-3).(-2xdx)\\ & =-3{\displaystyle \int {(1-x^2)}^4.(-2xdx)}\\ & =-3{\displaystyle \int {(1-x^2)}^4.d(1-x^2)}\\ & =-3\left[{\displaystyle \frac{{(1-x^2)}^5}{5}}\right]+C\\ & =-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \begin{array}{rl}{\displaystyle }& \int \frac{dx}{2022x+2023}=....\end{array}\\ & \begin{array}{l}\\ \text{a}.\ln |2022x+2023|+C\\ & \\ \text{b}.{\displaystyle \frac{1}{2021}\ln |2022x+2023|+C}\\ & \\ \text{c}.{\displaystyle \frac{1}{2022}\ln |2022x+2023|+C}\\ & \\ \text{d}.{\displaystyle \frac{1}{2023}\ln |2022x+2023|+C}\\ & \\ \text{e}.{\displaystyle \frac{2022}{2023}\ln |2022x+2023|+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int \frac{dx}{2022x+2023}}\\ & \text{Misalkan}\\ & u=2022x+2023\\ & \iff du=2022dx\iff {\displaystyle \frac{1}{2022}du=dx}\\ & \text{Sehingga}\\ & ={\displaystyle \int \frac{1}{u}.\left(\frac{1}{2022}du\right)={\displaystyle \frac{1}{2022}\int \frac{1}{u}du}}\\ & ={\displaystyle \frac{1}{2022}\ln |2022x+2023|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Diketahui}{\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\text{jika}f(4)=19,}\\ & \text{ maka}f(1)=....\\ & \begin{array}{l}\\ \text{a}.2\\ \text{b}.3\\ \text{c}.4\\ \text{d}.5\\ \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diingat bahwa}{\displaystyle \frac{d}{dx}f(x)=h(x)}\\ & \iff d(f(x))=h(x)dx\\ & \iff {\displaystyle \int d(f(x))={\displaystyle \int h(x)dx}}\\ & \text{Perhatikan kasus di pada soal di atas}\\ & {\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\iff d(f(x))=3\sqrt{x}dx}\\ & \iff {\displaystyle \int d(f(x))=\int 3\sqrt{x}dx=\int 3x^{{.}^{\frac{1}{2}}}dx}\\ & \iff f(x)=2x^{{.}^{\frac{3}{2}}}+C=2x\sqrt{x}+C\end{array}\\ & \begin{array}{rl}& \text{Karena diketahui}\\ & f(4)=19,\text{maka}\\ & \iff 2(4)(\sqrt{4})+C=19\\ & \iff 16+C=19\\ & \iff C=3\\ & \text{Sehingga}f(x)=2x\sqrt{x}+3\\ & \text{maka}f(1)=2.1.\sqrt{1}+3=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \begin{array}{rl}{\displaystyle }& -\int \frac{dx}{x^2+3x+2}dx=....\end{array}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \ln \left|{\displaystyle \frac{x+1}{x+2}}\right|+C}\\ & \\ \text{b}.{\displaystyle \ln \left|{\displaystyle \frac{x+2}{x+1}}\right|+C}\\ & \\ \text{c}.{\displaystyle \ln |x^2+3x+2|+C}\\ & \\ \text{d}.\arctan 2\left({\displaystyle x+\frac{3}{2}}\right)+C\\ & \\ \text{e}.-{\displaystyle \arctan 2\left({\displaystyle x+\frac{3}{2}}\right)+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle -\int \frac{dx}{x^2+3x+2}}& =-{\displaystyle \int \frac{1}{(x+1)(x+2)}dx}\\ & =-{\displaystyle \int \left({\displaystyle \frac{1}{x+1}-{\displaystyle \frac{1}{x+2}}}\right)dx}\\ & =-{\displaystyle \int \frac{1}{x+1}dx+{\displaystyle \int \frac{1}{x+2}dx}}\\ & ={\displaystyle \int \frac{1}{x+2}dx-{\displaystyle \int \frac{1}{x+1}dx}}\\ & ={\displaystyle \ln |x+2|-{\displaystyle \ln |x+1|+C}}\\ & ={\displaystyle \ln \left|{\displaystyle \frac{x+2}{x+1}}\right|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Tentukanah integral berikut}?\\ & \text{a}.\int dt\\ & \text{b}.\int 4dw\\ & \text{c}.\int (x^3+5)dx\\ & \text{d}.\int (x^{\frac{3}{2}}-2\sqrt{x}+1)dx\\ & \text{e}.\int (5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}})dx\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{c}\\ \begin{array}{r}\text{a}.\int dt=t+c\end{array}\\ \begin{array}{r}\text{b}.\int 4dw=4w+C\end{array}\\ \begin{array}{rl}\text{c}.\int (x^3+5)dx& ={\displaystyle \frac{1}{3+1}{x}^{3+1}+5x+C}\\ & ={\displaystyle \frac{1}{4}{x}^4+5x+C}\end{array}\end{array}\\ & \begin{array}{rl}\text{d}.& \int (x^{\frac{3}{2}}+2\sqrt{x}+1)dx\\ & =\int (x^{\frac{3}{2}}+2x^{\frac{1}{2}}+1)dx\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{3}{2}+1}}{x}^{\frac{3}{2}+1}+\frac{2}{{\displaystyle \frac{1}{2}+1}}{x}^{\frac{1}{2}+1}+x+C}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{5}{2}}}{x}^{\frac{5}{2}}+\frac{2}{{\displaystyle \frac{3}{2}}}{x}^{\frac{3}{2}}+x+C}\\ & =\frac{2}{5}{x}^{\frac{5}{2}}+\frac{4}{3}{x}^{\frac{3}{2}}+x+C\\ & \text{atau dapat juga kita menyatakan dengan}\\ & =\frac{2}{5}{x}^{2\frac{1}{2}}+\frac{4}{3}{x}^{1\frac{1}{2}}+x+C\\ & =\frac{2}{5}{x}^2\sqrt{x}+\frac{4}{3}x\sqrt{x}+x+C\end{array}\\ & \begin{array}{rl}\text{e}.& \int (5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}})dx\\ & =.....\\ & \text{silahkan dicoba sendiri sebagai latihan}\end{array}\end{array}$.

Contoh Soal 3 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 11.& {\displaystyle \int d\left(2x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x+C}\\ \text{b}.{\displaystyle 2x+C}\\ \text{c}.{\displaystyle 3x+C}\\ \text{d}.{\displaystyle 4x+C}\\ \text{e}.{\displaystyle 5x+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int d\left(2x\right)=\int 2dx=2x+C.}\end{array}$.

$\begin{array}{l}\\ 12.& {\displaystyle \int 5d\left(2x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 5+C}\\ \text{b}.{\displaystyle \frac{5}{2}x+C}\\ \text{c}.{\displaystyle 5x+C}\\ \text{d}.{\displaystyle 10x+C}\\ \text{e}.{\displaystyle 25x+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int 5d\left(2x\right)=\int 5.2dx=\int 10dx=10x+C.}\end{array}$.

$\begin{array}{l}\\ 13.& {\displaystyle \int 6xd\left(3x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18x^2+C}\\ \text{b}.{\displaystyle 9x^2+C}\\ \text{c}.{\displaystyle 5x^2+C}\\ \text{d}.{\displaystyle 3x^2+C}\\ \text{e}.{\displaystyle 2x^2+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int 6xd\left(3x\right)=\int 6x.3dx=\int 18xdx}\\ & ={\displaystyle \frac{18x^2}{2}+C=9x^2+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& {\displaystyle \int (8x^3-2x)d(-2x)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -16x^4+4x^2+C}\\ \text{b}.{\displaystyle 16x^4-4x^2+C}\\ \text{c}.{\displaystyle -4x^4-2x^2+C}\\ \text{d}.{\displaystyle -4x^4+2x^2+C}\\ \text{e}.{\displaystyle 4x^4-2x^2+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int (8x^3-2x)d(-2x)}\\ & =\int (8x^3-2x).(-2dx)\\ & =\int (-16x^3+4x)dx\\ & ={\displaystyle -\frac{16x^4}{4}+\frac{4x^2}{2}+C}\\ & ={\displaystyle -4x^4+2x^2+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& {\displaystyle \int (x+3)d(2x+6)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 4x^2+24x+C}\\ \text{b}.{\displaystyle 2x^2+12x+C}\\ \text{c}.{\displaystyle x^2+6x+C}\\ \text{d}.{\displaystyle 4x^2+C}\\ \text{e}.{\displaystyle x^2+C}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|cc|}\hline \text{Pertama}& \text{Kedua}\\ \begin{array}{rl}& {\displaystyle \int (x+3)d(2x+6)}\\ & ={\displaystyle \frac{1}{2}\int (2x+6)d(2x+6)}\\ & ={\displaystyle \frac{1}{2}\left[\frac{1}{2}{(2x+6)}^2\right]+C}\\ & ={\displaystyle \frac{1}{4}(4x^2+24x+36)+C}\\ & ={\displaystyle \frac{4x^2}{4}+\frac{24x}{4}+\frac{36}{4}+C}\\ & =x^2+6x+\underset{C}{\underbrace{9+C}}\\ & =x^2+6x+C\end{array}& \begin{array}{rl}& {\displaystyle \int (x+3)d(2x+6)}\\ & =\int (x+3).\left(2dx\right)\\ & =\int (2x+6)dx\\ & ={\displaystyle \frac{2x^2}{2}+6x+C}\\ & =x^2+6x+C\\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$ .

Contoh Soal 2 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 6.& {\displaystyle \int x^3\sqrt{x}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}\\ \text{b}.{\displaystyle \frac{9}{2}{x}^4\sqrt{x}+C}\\ \text{c}.{\displaystyle \frac{1}{9}{x}^4\sqrt{x}+C}\\ \text{d}.{\displaystyle 9x^4\sqrt{x}+C}\\ \text{e}.{\displaystyle x^4\sqrt{x}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{x}dx=\int x^3.x^{\frac{1}{2}}dx}\\ & =\int x^{3\frac{1}{2}}dx=\int x^{\frac{7}{2}}dx={\displaystyle \frac{x^{\frac{7}{2}+1}}{{\displaystyle \frac{7}{2}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{9}{2}}}{{\displaystyle \frac{9}{2}}}+C={\displaystyle \frac{2}{9}{x}^{4\frac{1}{2}}+C={\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{17}{6}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{b}.{\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{c}.{\displaystyle x^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{d}.{\displaystyle \frac{6}{17}x\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{e}.{\displaystyle \frac{1}{2}x\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx}\\ & ={\displaystyle \int x{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & ={\displaystyle \int x^{1+\frac{1}{2}+\frac{2}{6}}dx}\\ & ={\displaystyle \int x^{\frac{11}{6}}dx={\displaystyle \frac{x^{\frac{11}{6}+1}}{{\displaystyle \frac{11}{6}+1}}+C}}\\ & ={\displaystyle \frac{x^{\frac{17}{6}}}{{\displaystyle \frac{17}{6}}}+C={\displaystyle \frac{6}{17}{x}^2.x^{\frac{5}{6}}+C}}\\ & ={\displaystyle \frac{6}{17}{x}^2{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& {\displaystyle \int x^2\sqrt{x^2\sqrt[3]{x^3}}dx=....}\\ & \begin{array}{l}\\ \mathbf{\text{a}}.{\displaystyle x^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{b}}.{\displaystyle \frac{27}{6}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{c}}.{\displaystyle \frac{6}{27}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{d}}.{\displaystyle \frac{6}{21}{x}^2\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{e}}.{\displaystyle \frac{6}{21}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^2\sqrt{x^2\sqrt[3]{x^3}}dx=\int x^2{(x^2.x^1)}^{\frac{1}{2}}dx}\\ & =\int x^{2+\frac{2}{2}+\frac{1}{2}}dx=\int x^{\frac{7}{2}}dx={\displaystyle \frac{x^{\frac{7}{2}+1}}{{\displaystyle \frac{7}{2}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{9}{2}}}{{\displaystyle \frac{9}{2}}}+C}\\ & ={\displaystyle \frac{2}{9}{x}^{\frac{9}{2}}+C}\\ & ={\displaystyle \frac{2.3}{9.3}{x}^3.x^{\frac{3}{2}}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3{\left(x^3\right)}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3\sqrt{x^2.x}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& {\displaystyle \int x^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{b}.{\displaystyle \frac{6}{23}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{c}.{\displaystyle \frac{23}{6}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{d}.{\displaystyle \frac{23}{6}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{\frac{1}{x}\sqrt[3]{x^2}}dx=\int x^3{(x^{-1}.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & =\int x^{3-\frac{1}{2}+\frac{1}{3}}dx=\int x^{\frac{17}{6}}dx\\ & ={\displaystyle \frac{x^{\frac{17}{6}+1}}{{\displaystyle \frac{17}{6}+1}}+C={\displaystyle \frac{x^{\frac{23}{6}}}{{\displaystyle \frac{23}{6}}}+C}}\\ & ={\displaystyle \frac{6}{23}{x}^{\frac{24-1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.x^{-\frac{1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.{\left(x^{-\frac{1}{3}}\right)}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{x^{-1+\frac{2}{3}}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{\frac{1}{x}\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& {\displaystyle \int x\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{b}.{\displaystyle \frac{13}{8}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{c}.{\displaystyle \frac{8}{13}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{d}.{\displaystyle \frac{1}{2}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{e}.{\displaystyle \frac{8}{5}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}dx}\\ & =\int x{\left(x^{-1}{\left(x{\left(x^{-1}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}dx\\ & =\int x(x^{-\frac{1}{2}}.x^{\frac{1}{4}}.x^{-\frac{1}{8}})dx\\ & =\int x^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}}dx=\int x^{\frac{5}{8}}dx\\ & ={\displaystyle \frac{x^{\frac{5}{8}+1}}{{\displaystyle \frac{5}{8}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{13}{8}}}{{\displaystyle \frac{13}{8}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^{\frac{16-3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2.x^{-\frac{3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-\frac{3}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-1+\frac{1}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}.{\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x^{1-\frac{1}{2}}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C}\end{array}\end{array}$

Contoh Soal 1 Materi Integral Tak Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& {\displaystyle \int x^{2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^3+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^6+C}\\ \text{c}.{\displaystyle \frac{1}{3}{x}^6+C}\\ \text{d}.{\displaystyle \frac{1}{6}{x}^3+C}\\ \text{e}.{\displaystyle \frac{2}{3}{x}^3+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{2}dx={\displaystyle \frac{x^{2+1}}{2+1}+C={\displaystyle \frac{1}{3}{x}^3+C}}}\end{array}$

$\begin{array}{l}\\ 2.& {\displaystyle \int x^{-2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2x^{-1}+C}\\ \text{b}.{\displaystyle -x^{-1}+C}\\ \text{c}.{\displaystyle -\frac{1}{2}{x}^{-2}+C}\\ \text{d}.{\displaystyle -\frac{1}{3}{x}^{-3}+C}\\ \text{e}.{\displaystyle -3x^{-3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{-2}dx={\displaystyle \frac{x^{-2+1}}{-2+1}+C={\displaystyle \frac{1}{-1}{x}^{-1}+C=-x^{-1}+C}}}\end{array}$.

$\begin{array}{l}\\ 3.& {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{4}{x}^{\frac{4}{3}}+C}\\ \text{b}.{\displaystyle x^{\frac{4}{3}}+C}\\ \text{c}.{\displaystyle \frac{3}{4}{x}^{\frac{2}{3}}+C}\\ \text{d}.{\displaystyle x^{-\frac{2}{3}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^{-\frac{2}{3}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx={\displaystyle \frac{x^{\frac{1}{3}+1}}{{\displaystyle \frac{1}{3}+1}}+C={\displaystyle \frac{1}{{\displaystyle \frac{4}{3}}}{x}^{{.}^{\frac{4}{3}}}+C={\displaystyle \frac{3}{4}{x}^{{.}^{\frac{4}{3}}}+C.}}}}\end{array}$.

$\begin{array}{l}\\ 4.& {\displaystyle \int \frac{1}{x^3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -\frac{1}{2x^2}+C}\\ \text{b}.{\displaystyle -\frac{2}{x^2}+C}\\ \text{c}.{\displaystyle \frac{1}{3x^4}+C}\\ \text{d}.{\displaystyle \frac{3}{x^4}+C}\\ \text{e}.{\displaystyle -\frac{1}{4x^3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{x^3}dx=\int x^{-3}dx}\\ & ={\displaystyle \frac{x^{-3+1}}{-3+1}+C=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& {\displaystyle \int \frac{1}{3}{x}^{3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^4+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^4+C}\\ \text{c}.{\displaystyle x^4+C}\\ \text{d}.{\displaystyle \frac{1}{12}{x}^4+C}\\ \text{e}.{\displaystyle \frac{4}{3}{x}^4+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int \frac{1}{3}{x}^{3}dx={\displaystyle \frac{1}{3}.\frac{x^{3+1}}{3+1}+C={\displaystyle \frac{1}{12}{x}^4+C}}}\end{array}$.

Penggunaan Integral Tak Tentu

Penggunaan Integral Tak Tentu

Penggunaan integral tak tentu ini dapat digunakan dalam menentukan suatu fungsi jika turunan dari fungsi tersebut diberikan. Selain itu untuk menentukan posisi, kecepatan, percepatan suatu benda pada waktu tertentu.

$\begin{array}{rl}& \text{Misalkan}\\ & \bullet s\text{adalah menunjukkan posisi}\\ & \bullet v\text{adalah menunjukkan kecepatan}\\ & \bullet a\text{adalah menunjukkan percepatan}\\ & \bullet t\text{adalah menunjukkan waktu}\\ \\ & \mathbf{\text{Perhatikan hubungan berikut}}\\ & v={\displaystyle \frac{ds}{dt}\iff ds=vdt}\\ & \iff {\displaystyle \int ds=\int vdt}\\ & \iff s={\displaystyle \int vdt}\\ \\ & \mathbf{\text{Demikian juga hubungan berikut}}\\ & a={\displaystyle \frac{dv}{dt}\iff dv=adt}\\ & \iff {\displaystyle \int dv=\int adt}\\ & \iff v={\displaystyle \int adt}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}y,\text{Jika}{\displaystyle \frac{dy}{dx}=2022x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{dy}{dx}=2022x\iff dy=2022xdx}\\ & \iff {\displaystyle \int dy=\int 2022xdx}\\ & \iff y={\displaystyle \frac{2022}{2}{x}^2+C}\\ & \iff y=1011x^2+C\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi turunan pertama kurva}\\ & \text{adalah}{\displaystyle \frac{dy}{dx}=2x-2.\text{Jika kurva melalui}}\\ & \text{titik}(3,2),\text{tentukan persamaan dari kurva}\\ & \text{tersebut}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & {\displaystyle \frac{dy}{dx}=2x-2}\\ & \iff dy=(2x-2)dx\\ & \iff {\displaystyle \int dy=\int (2x-2)dx}\\ & \iff y=x^2-2x+C\\ & \text{Karena kurva melalui}(3,2),\text{maka}\\ & (2)=(3{)}^2-2(3)+C\iff 2=3+C\\ & \iff C=-1\\ & \text{Jadi},y=x^2-2x-1\text{atau}\\ & f(x)=x^2-2x-1\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui bahwa}{f}^{″}(x)=x^2.\text{Jika}{f}^{\prime }(0)=6\\ & \text{dan}f(0)=3,\text{tentukanlah}f(x)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & {\displaystyle f^{″}(x)=x^2}\\ & \iff {\displaystyle \int f^{″}(x)dx={\displaystyle \int x^{2}dx}}\\ & \iff {\displaystyle f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C}}\\ & \text{Karena}{f}^{\prime }(x)=6,\text{maka}\\ & f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C\iff 6={\displaystyle \frac{1}{3}(0{)}^3+C}}\\ & \iff C=6\\ & \text{Sehingga}{f}^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+6}\\ & \text{Selanjutnya}{\displaystyle \int f^{\prime }(x)dx={\displaystyle \int {\displaystyle \frac{1}{3}{x}^3+6dx}}}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+C}\\ & \text{Dan juga karena}f(0)=3,\text{maka}\\ & f(0)={\displaystyle \frac{1}{12}(0{)}^4+6(0)+C=3}\\ & C=3,\text{sehingga diperoleh}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\\ & \text{Jadi},f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}{f}^{\prime }\left(x\right)=6x^2-2x+6\text{dan}\\ & \text{nilai}\text{fungsi}f\left(2\right)=-7.\text{Tentukanlah}\\ & \text{rumus}\text{fungsi}\text{tersebut}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f\left(x\right)& =\int f^{\prime }\left(x\right)dx\\ & =\int (6x^2-2x+6)dx\\ & =2x^3-x^2+6x+C\end{array}\\ \\ & \text{Karena}f\left(2\right)=-7,\text{maka}\\ \\ & \begin{array}{rl}f\left(2\right)& ={2.2}^3-2^2+6.2+C\\ \iff -7& =16-4+12+C\\ \iff C& =-31\end{array}\\ \\ & \text{Jadi},f\left(x\right)=2x^3-x^2+6x-31\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}f(x),\text{Jika diketahui}\\ & \text{a}.f^{\prime }(x)=12x^3\text{dan}f(1)=8\\ & \text{b}.f^{\prime }(x)=x^2-2x+3\text{dan}f(3)=9\\ & \text{c}.f^{″}(x)=2,f^{\prime }(2)=2\text{dan}f(2)=10\\ & \text{d}.f^{″}(x)=x^2,f^{\prime }(0)=6\text{dan}f(0)=3\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan persamaan kurva}f(x)\\ & \text{disetiap titik}(x,y)\text{yang memenuhi}\\ & \text{syarat berikut}\\ & \text{a}.{\displaystyle \frac{dy}{dx}=4x+1\text{dan kurva melalui}(0,2)}\\ & \text{b}.{\displaystyle \frac{dy}{dx}={\displaystyle \frac{1}{x^2}+3\text{dan kurva melalui}(1,4)}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}2x-y=3\text{merupakan garis}\\ & \text{kurva di titik}(1,-1).\text{Jika di tiap titik}\\ & \text{pada kurva berlaku}{\displaystyle y^{″}=2x^2-3x+1}\\ & \text{tentukan persamaan kurva tersebut}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui kecepatan sebuah sepeda}\\ & \text{diformulasian dengan}{\displaystyle \frac{ds}{dt}=3t^2+4t-1}\\ & \text{Jika}s\text{menyatakan jarak yang ditempuh}\\ & \text{dalam satuan meter},t\text{menyatakan waktu}\\ & \text{dengan}s=5\text{untuk}t=2,\text{tentukanlah}\\ & \text{jarak yang ditempuh pengendara dalam}\\ & \text{waktu 5 menit}\end{array}$.

Teknik Pengintegralan (Bagian 2)

2. Integral Parsial

2. 1 Integral Parsial

Jika teknik pada no.1 pada pembahasan sebelumnya tidak dapat digunakan, maka kemungkinan adalah dengan menggunakan teknik yang satunya ini, yaitu teknik integral parsial. Adapun untuk teknik integral ini diformulasikan dengan bentuk rumus

${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Perhatian kembali soal berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{Dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\\ \\ & \text{Alternatif 2}\\ & \begin{array}{l}\\ {\displaystyle \int x\sqrt{x-1}dx=\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}}& \\ \begin{array}{cccc}u=x& & & dv=\sqrt{x-1}dx\\ du=dx& & & {\displaystyle \int dv=\int \sqrt{x-1}dx}\\ & & & v=\int {(x-1)}^{\frac{1}{2}}dx\\ & & & v={\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\end{array}\end{array}\\ & \text{Dengan integral parsial}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}=u.v-\int v.du\\ & =x.\left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)-\int \left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)dx\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{2}{3}\times \frac{2}{5}{(x-1)}^{\frac{5}{2}}+C}}\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\end{array}\end{array}\end{array}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & \begin{array}{|cc|}\hline \text{Hasil dengan Substitusi}& \text{Hasil dengan Integral Parsial}\\ {\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& {\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ \hline\end{array}\\ & \text{Jika kita sejajarkan dengan ruas }\\ & \text{yang berbeda, maka}\\ & \begin{array}{rl}{\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ {\displaystyle \frac{2}{5}(x-1){(x-1)}^{\frac{3}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}(x-1){(x-1)}^{\frac{3}{2}}}}\\ \left({\displaystyle \frac{2}{5}(x-1)+\frac{2}{3}}\right){(x-1)}^{\frac{3}{2}}& =\left({\displaystyle \frac{2x}{3}-\frac{4}{15}(x-1)}\right){(x-1)}^{\frac{3}{2}}\\ {\displaystyle \frac{2x}{5}-\frac{2}{5}+\frac{2}{3}}& ={\displaystyle \frac{2x}{3}-\frac{4x}{15}+\frac{4}{15}}\\ {\displaystyle \frac{2x}{5}+\frac{4}{15}}& ={\displaystyle \frac{2x}{5}+\frac{4}{15}}\\ \text{ruas kiri}& =\text{ruas kanan}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah integral dari}\int (x+2)dx\\ & \text{dengan cara}\\ & \text{a})\text{substitusi}\\ & \text{b})\text{parsial}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \mathbf{\text{Cara substitusi}}\\ & \int (x+2)dx=........?\\ & \text{Misalkan}m=x+2\\ & dm=dx\\ & \text{maka},\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{m}}\end{array}}{\underbrace{(x+2)}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dm}}\end{array}}{\underbrace{dx}}\\ & =\frac{1}{2}{m}^2+C=\frac{1}{2}(x+2{)}^2+C\\ & =\frac{1}{2}(x^2+4x+4)+C=\frac{1}{2}{x}^2+2x+\underset{\begin{array}{c}|\\ \mathbf{\text{C}}\end{array}}{\underbrace{2+C}}\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\\ \\ & \begin{array}{rl}& \mathbf{\text{Cara parsial}}\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{u}}\end{array}}{\underbrace{1}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dv}}\end{array}}{\underbrace{(x+2)dx}}\\ & \{\begin{array}{c}u=1\to du=0\\ \\ v=\underset{\begin{array}{c}|\\ =\frac{1}{2}{x}^2+2x+C\end{array}}{\int dv}\leftarrow dv=(x+2)dx\end{array}\\ & =\mathbf{\text{u.v}}-\int \mathbf{\text{v.du}}\\ & =1.(\frac{1}{2}{x}^2+2x+C)-\int (\frac{1}{2}{x}^2+2x+C).0\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\end{array}$.

2. 2 Aturan Tanzalin

Sumber Referensi

$\begin{array}{rl}& \text{Tentukanlah hasil integral berikut}\\ & {\displaystyle \int 2x^2(x-4{)}^{2}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Bentuk}{\displaystyle \int 2x^2(x-4{)}^{2}dx\text{dianggap sebagai}}\\ & {\displaystyle \int udv\text{dengan}u\text{adalah bagian yang mudah}}\\ & \text{kita diferensialkan, maka}u\text{kita pilihkan}\\ & \text{yaitu}u=2x^2\\ & \text{Selanjutnya dengan}\mathbf{\text{aturan Tanzalin}}\\ & \text{sebagai berikut}\end{array}$.

$\begin{array}{|cc|}\hline \text{Diderensialkan}& \text{Diintegralkan}\\ \begin{array}{rl}& +2x^2\\ & -4x\\ & +4\\ & -0\\ & \end{array}& \begin{array}{rl}& (x-4{)}^4\\ & {\displaystyle \frac{1}{5}(x-4{)}^5}\\ & {\displaystyle \frac{1}{30}(x-4{)}^6}\\ & {\displaystyle \frac{1}{210}(x-4{)}^7}\end{array}\\ \hline\end{array}$.

Hasil dari integral teknik ini adalah:

$\begin{array}{rl}& {\displaystyle \int 2x^2(x-4{)}^{4}dx}\\ & =(+2x^2)\left({\displaystyle \frac{1}{5}(x-4{)}^5}\right)+(-4x)\left({\displaystyle \frac{1}{30}{(x-4)}^6}\right)\\ & +(+4)\left({\displaystyle \frac{1}{210}(x-4{)}^7}\right)+C\\ & ={\displaystyle \frac{2}{5}{x}^2(x-4{)}^5-{\displaystyle \frac{2}{15}x(x-4{)}^6+{\displaystyle \frac{2}{105}(x-4{)}^7+C}}}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{1. Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle x^2\sqrt{x+7}dx}}& \text{d}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int 2x^2\sqrt{3-x}dx}& \text{e}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int 3x^2\sqrt{2x+1}dx}& \text{f}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\end{array}\end{array}$ .

$\begin{array}{rl}& \text{2. Selesaikan soal berikut dengan Metode Tanzalin}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\\ & \text{d}.& {\displaystyle \int {\displaystyle 3(x-2{)}^4.\sqrt[3]{x}dx}}\\ & \text{e}.& {\displaystyle \int {\displaystyle \frac{(x^4-3)}{\sqrt{1-x}}dx}}\\ & \text{f}.& {\displaystyle \int {\displaystyle x^3\sqrt{1-2x}dx}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Teknik Pengintegralan (Bagian 1)

1. Integral Substitusi

Ada beberapa bentuk integral yang terkadang pengintegralannya membutuhkan teknik tertentu. Di antara bentuk tertentu itu adalah dengan substitusi, yaitu:

${\displaystyle \int u^n.u^{\prime }dx={\displaystyle \int u^{n}du={\displaystyle \frac{1}{n+1}{u}^{n+1}+C}}}$ .

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^2+3)}^{20}2xdx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^2& +& 3& ,& \text{maka}\\ du& =& 2x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {(x^2+3)}^{20}2xdx=\int u^{20}du\\ & =\frac{1}{21}{u}^{21}+C=\frac{1}{21}{(x^2+3)}^{21}+C\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^4-x^2)}^5(16x^3-8x)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^4& -& x^2& ,& \text{maka}\\ du& =& 4x^3& -& 2x& dx\\ 4du& =& 16x^3& -& 8x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int u^5.4du=\frac{4}{6}{u}^6+C}\\ & ={\displaystyle \frac{2}{3}{(x^4-x^2)}^6+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{x+2}{x^2+4x+4}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccccc}u& =& x^2& +& 4x& +& 4& ,& \text{maka}\\ du& =& 2x& +& 4& dx\\ \frac{1}{2}du& =& x& +& 2& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{u}.\frac{1}{2}du={\displaystyle \frac{1}{2}\int \frac{1}{u}du}}\\ & ={\displaystyle \frac{1}{2}\ln u+C=\frac{1}{2}\ln (x^2+4x+4)+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{e^{3y}}{{(1-2e^{3y})}^2}dy}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& 1& -& 2e^{3y}& ,& \text{maka}\\ du& =& -& 6e^{3y}& dy& ,\\ -\frac{1}{6}du& =& e^{3y}& dy\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int -\frac{1}{6}\frac{1}{u^2}du=-\frac{1}{6}\int \frac{du}{u^2}}\\ & =-{\displaystyle \frac{1}{6}(-1)\left(u^{-1}\right)+C}\\ & ={\displaystyle \frac{1}{6u}+C}\\ & ={\displaystyle \frac{1}{6}.\frac{1}{1-2e^{3y}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{r}\text{Misalkan}\\ u& =x^2+3\\ du& =2xdx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ & =\int {(x^2+3)}^5.6.2xdx\\ & =6\int \underset{u^5}{\underbrace{{(x^2+3)}^5}}.\underset{du}{\underbrace{2xdx}}\\ & =6\int u^{5}du\\ & =6.{\displaystyle \frac{u^{5+1}}{5+1}+C}\\ & =u^6+C\\ & ={(x^2+3)}^6+C\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle \frac{x^5}{x^6+a}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^6+a\\ & \iff du=6x^{5}dx\\ & \iff {\displaystyle \frac{1}{6}du=x^{5}dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {\displaystyle \frac{x^5}{x^6+a}dx}\\ & =\int {\displaystyle \frac{1}{x^6+a}.x^{5}dx}\\ & =\int {\displaystyle \frac{1}{\underset{u}{\underbrace{x^6+a}}}.\underset{\frac{1}{6}du}{\underbrace{x^{5}dx}}}\\ & ={\displaystyle \frac{1}{6}\int \frac{1}{u}du}\\ & ={\displaystyle \frac{1}{6}\ln \left|u\right|+C}\\ & ={\displaystyle \frac{1}{6}\ln |x^6+a|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle (x^2+1){(x^3+3x+2)}^{5}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^3+3x+2\\ & \iff du=(3x^2+3)dx\\ & \iff du=3(x^2+1)dx\\ & \iff {\displaystyle \frac{1}{3}du=(x^2+1)dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}\int & (x^2+1){(x^3+3x+2)}^{5}dx\\ & =\int {(x^3+3x+2)}^5.(x^2+1)dx\\ & =\int \underset{u^5}{\underbrace{{(x^3+3x+2)}^5}}.\underset{\frac{1}{3}du}{\underbrace{(x^2+1)dx}}\\ & ={\displaystyle \frac{1}{3}\int u^{5}du}\\ & ={\displaystyle \frac{1}{3}.{\displaystyle \frac{u^{5+1}}{5+1}+C}}\\ & ={\displaystyle \frac{1}{18}.u^6+C}\\ & ={\displaystyle \frac{1}{18}{(x^3+3x+2)}^6+C}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle \frac{4x^6+3x^5-8}{x^5}dx}}& \text{f}.& {\displaystyle \int {\displaystyle \frac{{(\sqrt{x}+4)}^3}{\sqrt{x}}dx}}\\ & \text{b}.& {\displaystyle \int {\displaystyle \frac{x^2}{\sqrt{4-x^3}}dx}}& \text{g}.& {\displaystyle \int {\displaystyle \frac{x+3}{\sqrt[5]{{(x^2+6x-1)}^2}}dx}}\\ & \text{c}.& {\displaystyle \int x^3{(x^4+10)}^{{.}^{-\frac{2}{3}}}dx}& \text{h}.& {\displaystyle \int x^5\sqrt{x^6+1}dx}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Integral Fungsi Aljabar

A. Pengertian

Pengintegralan dari suatu fungsi $f(x)$ berbentuk  $\int f(x)dx$ dapat disebut sebagai integral tak tentu dari fungsi  $f(x)$ dan jika  $F(x)$ adalah anti turunan dari  $f(x)$, maka  $F(x)dx=F(x)+C$.

$\begin{array}{rl}& \text{Dengan}\\ & \begin{array}{rl}F(x)& =\text{fungsi integral dari}f(x)\\ f(x)& =\text{fungsi yang diintegralkan}\\ C& =\mathbf{\text{Konstanta}}\end{array}\end{array}$.

B. Rumus Dasar Integral tak Tentu Fungsi Ajabar

Berikut rumus dasar yang perlu diingat

$\begin{array}{rl}\bullet & \int dx=x+C\\ \bullet & \int k.\left({\displaystyle f(x)}\right)dx=k\int f(x)dx\\ \bullet & \int (f(x)\pm g(x))dx=\int f(x)dx+\int g(x)dx\\ \bullet & \int a{x}^{n}dx={\displaystyle \frac{a}{n+1}{x}^{n+1}+C}\end{array}$.

Sebagai rumus-rumus integral yang lain adalah sebagai berikut

$\begin{array}{l}\\ \bullet & {\displaystyle \int a{x}^{n}dx=\frac{a}{n+1}.x^{n+1}+C,\text{dengan}n\ne -1}\\ \bullet & {\displaystyle \int adx=ax+C}\\ \bullet & {\displaystyle \int \frac{1}{x}dx=\int x^{-1}dx=\ln x+C}\\ \bullet & {\displaystyle \int \left|x\right|dx=\frac{1}{2}x\left|x\right|+C}\\ \bullet & {\displaystyle \int \ln xdx=x\ln x-x+C}\\ \bullet & {\displaystyle \int e^{x}dx=e^x+C}\\ \bullet & {\displaystyle \int a^{x}dx=\frac{a^x}{\ln a}+C}\\ \bullet & {\displaystyle \int e^{ax}dx=\frac{1}{a}.e^{ax}+C}\\ \bullet & {\displaystyle \int (x^m+x^n+...+x^p)dx}\\ & ={\displaystyle \int x^{m}dx+\int x^{n}dx+...+\int x^{p}d}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int x^{5}dx}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{1}{x^{2022}}dx}}\\ & \text{d}.{\displaystyle \int 2022ydy}\\ & \text{e}.{\displaystyle \int e^{2022x}dx}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int x^{5}dx=\frac{1}{5+1}.x^{5+1}+C=\frac{1}{6}.x^6+C}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx=\frac{2022}{5+1}.x^{5+1}+C=337x^6+C}\\ & \text{c}.{\displaystyle \int \frac{1}{x^{2022}}dx=\int x^{-2022}dx}\\ & ={\displaystyle \frac{1}{-2022+1}.x^{-2022+1}+C=-\frac{1}{2021}.x^{-2021}+C}\\ & =-{\displaystyle \frac{1}{2021x^{2021}}+C}\\ & \text{d}.{\displaystyle \int 2022ydy=\frac{2022}{1+1}{y}^{1+1}+C=1011y^2+C}\\ & \text{e}.{\displaystyle \int e^{2022x}dx=\frac{1}{2022}.e^{2022x}+C}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx=\frac{{2022}^x}{\ln 2022}+C}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx}\\ & \text{b}.{\displaystyle \int (x^2-2xy+y^2)dx}\\ & \text{c}.{\displaystyle \int |x-1|+|x-2|dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.{\displaystyle }& \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx\\ & ={\displaystyle \int 3x^{2}dx-\int xdx+2\int dx-\int \frac{1}{x}dx+3\int \frac{1}{x^2}dx}\\ & ={\displaystyle \frac{3}{2+1}{x}^{2+1}-\frac{1}{1+1}{x}^{1+1}+2x-\ln x+3\left(\frac{1}{-2+1}{x}^{-2+1}\right)+C}\\ & ={\displaystyle \frac{2}{3}{x}^3-\frac{1}{2}{x}^2+2x-\ln x-\frac{3}{x}+C}\end{array}\\ & \begin{array}{rl}\text{b}.{\displaystyle }& \int (x^2-2xy+y^2)dx\\ & ={\displaystyle \int x^{2}dx-2y\int xdx+y^2\int dx}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-\frac{2y}{1+1}{x}^{1+1}+y^2.x+C}\\ & ={\displaystyle \frac{1}{3}{x}^3-x^{2}y+x{y}^2+C}\end{array}\\ & \begin{array}{rl}\text{c}.{\displaystyle }& \int |x-1|+|x-2|dx\\ & ={\displaystyle \frac{(x-1)}{2}|x-1|+\frac{(x-2)}{2}|x-2|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx={\displaystyle \frac{2}{{\displaystyle \frac{2}{3}+1}}{x}^{{.}^{\frac{2}{3}+1}}+C={\displaystyle \frac{6}{5}{x}^{{.}^{\frac{5}{3}}}+C}}}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx={\displaystyle \frac{1}{3}{x}^{{.}^{\frac{3}{4}}}dx=\left({\displaystyle \frac{1}{3}}\right).{\displaystyle \frac{1}{{\displaystyle \frac{3}{4}+1}}{x}^{{.}^{\frac{3}{4}+1}}+C}}}\\ & =\left({\displaystyle \frac{1}{3}}\right){\displaystyle \frac{4}{7}{x}^{{.}^{\frac{7}{4}}}+C={\displaystyle \frac{4}{21}{x}^{{.}^{\frac{7}{4}}}+C}}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx=\int x^2-4xdx}}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-{\displaystyle \frac{4}{1+1}{x}^{1+1}+C}}\\ & ={\displaystyle \frac{1}{3}{x}^3-2x^2+C}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx={\displaystyle \int {\displaystyle \frac{x^{{.}^{\frac{1}{2}}}(6x^2-2x)}{x}dx}}}}\\ & ={\displaystyle \int {\displaystyle \frac{6x^{{.}^{\frac{5}{2}}}-2x^{{.}^{\frac{3}{2}}}}{x}dx={\displaystyle \int 6x^{{.}^{\frac{3}{2}}}-2x^{{.}^{\frac{1}{2}}}dx}}}\\ & ={\displaystyle \frac{6}{{\displaystyle \frac{3}{2}+1}}{x}^{{.}^{\frac{3}{2}+1}}-\frac{2}{{\displaystyle \frac{1}{2}+1}}{x}^{{.}^{\frac{1}{2}+1}}+C}\\ & ={\displaystyle \frac{12}{5}{x}^{{.}^{\frac{5}{2}}}-{\displaystyle \frac{4}{3}{x}^{{.}^{\frac{3}{2}}}+C={\displaystyle \frac{12}{5}{x}^2\sqrt{x}-{\displaystyle \frac{4}{3}x\sqrt{x}+C}}}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ & ={\displaystyle \int (24x^3+4x^2-4x)dx}\\ & ={\displaystyle \frac{24}{3+1}{x}^{3+1}+\frac{4}{2+1}{x}^{2+1}-\frac{4}{1+1}{x}^{1+1}dx}\\ & =6x^4+{\displaystyle \frac{4}{3}{x}^3-2x^2+C}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2022dx}\\ & \text{b}.{\displaystyle \int \frac{dx}{2022}}\\ & \text{c}.{\displaystyle \int 2022xdx}\\ & \text{d}.{\displaystyle \int -2022x^{2}dx}\\ & \text{e}.{\displaystyle \int (x+2022)dx}\\ & \text{f}.{\displaystyle \int (-2022x^3+2023x^2-2024)dx}\\ & \text{g}.{\displaystyle \int x\sqrt{x}dx}\\ & \text{h}.{\displaystyle \int \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x}}}}}dx}\\ & \text{i}.{\displaystyle \int \frac{2022}{\sqrt[3]{x^2}}dx}\\ & \text{j}.{\displaystyle \int \frac{2022x}{\sqrt[3]{x^5}}dx}\\ & \text{k}.{\displaystyle \int (2022-2020t+t^2)dt}\\ & \text{l}.{\displaystyle \int (\frac{3}{t^3}+\frac{2}{t^2}+2022)dt}\\ & \text{m}.{\displaystyle \int (\sqrt{t}+\frac{1}{2\sqrt{t}})dt}\\ & \text{n}.{\displaystyle \int (a{y}^4+b{y}^2)dy}\\ & \text{o}.{\displaystyle \int (4a{x}^3+3b{x}^2+2cx+1)dx}\\ & \text{p}.{\displaystyle \int \frac{x^2+2022}{x^2}dx}\\ & \text{q}.{\displaystyle \int (e^x+e^{-x})dx}\\ & \text{r}.{\displaystyle \int e^{2023x}dx}\\ & \text{s}.{\displaystyle \int \frac{dx}{e^{2023x}}dx}\\ & \text{t}.{\displaystyle \int \left(\sqrt{{10}^x}\right)dx}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (2022x-2202)dx}\\ & \text{b}.{\displaystyle \int (x^2-2x-8)dx}\\ & \text{c}.{\displaystyle \int \sqrt{2x}dx}\\ & \text{d}.{\displaystyle \int x^2(x+2)(x-1)dx}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 3x(2x-{\displaystyle \frac{1}{x}})dx}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{(x^2-4)}{x^2}dx}}\\ & \text{c}.{\displaystyle \int {(2x-{\displaystyle \frac{1}{x^2}})}^{2}dx}\\ & \text{d}.{\displaystyle \int x^2\left({\displaystyle \frac{(x+4)(x-3)}{x}}\right)dx}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{2x^3-3x}{\sqrt{x}}dx}}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{2}{\sqrt{x}}{(1-\sqrt{x^2})}^{2}dx}}\\ & \text{c}.{\displaystyle \int x^2{(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}})}^{2}dx}\\ & \text{d}.{\displaystyle \int {\displaystyle \frac{x^2{(2+\sqrt[3]{x^4})}^2}{\sqrt{x}}dx}}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.