Contoh Soal 2 Materi Integral Tak Tentu Fungsi Aljabar

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

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

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

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

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