Q112 / 160 · TS EAMCET 2016 · Mathematics
$\int \frac{x^3}{\sqrt{1+x^2}} d x$ is equal to
- A$\sqrt{1+x^2}-\frac{x}{3}\left(1+x^2\right)^{3 / 2}+C$
- B$x \sqrt{1+x^2}+\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C$
- Correct answerC$x^2 \sqrt{1+x^2}-\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C$
- D$x^2 \sqrt{1+x^2}-\frac{1}{3}\left(1+x^2\right)^{1 / 2}+C$
Solution
$$
\begin{aligned}
& \text { Let } I=\int \frac{x^3}{\sqrt{1+x^2}} d x \\
& =\int x^2 \cdot \frac{x}{\sqrt{1+x^2}} d x \\
&
\end{aligned}
$$
By using integration by parts
$\begin{aligned} & =x^2 \int \frac{x}{\sqrt{1+x^2}} d x-\int\left\{2 x \int \frac{x}{\sqrt{1+x^2}} d x\right\} d x \\ & =x^2 \sqrt{1+x^2}-\int 2 x \sqrt{1+x^2} d x \\ & =x^2 \sqrt{1+x^2}-\frac{\left(1+x^2\right)^{3 / 2}}{\frac{3}{2}}+C \\ & =x^2 \sqrt{1+x^2}-\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C\end{aligned}$
Hence, the correct answer is option: 'C'.