Contoh Soal 4 Materi Integral Tak Tentu Fungsi Aljabar

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

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

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

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

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