Q56 / 120 · VITEEE 2008 · Mathematics
If $I=\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x$, then I is equal to
- A$\frac{2}{9}\left(1+x^{3}\right)^{\frac{5}{2}}+\frac{2}{3}\left(1+x^{3}\right)^{\frac{3}{2}}+C$
- B$\log \left|\sqrt{x}+\sqrt{1+x^{3}}\right|+C$
- C$\log \left|\sqrt{x}-\sqrt{1+x^{3}}\right|+C$
- Correct answerD$\frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+C$
Solution
$\mathrm{I}=\int \frac{\mathrm{x}^{5}}{\sqrt{1+\mathrm{x}^{3}}} \mathrm{dx}=\int \frac{\mathrm{x}^{3} \cdot \mathrm{x}^{2}}{\sqrt{1+\mathrm{x}^{3}}} \mathrm{dx}$
Let $1+x^{3}=t^{2}$, so that $3 x^{2} d x=2 t d t$
$\begin{array}{l}
\Rightarrow x^{2} d x=\frac{2}{3} t d t \\
\left.\therefore I=\int \frac{\left(t^{2}-1\right) \frac{2}{3} t d t}{t}=\frac{2}{3} \int t^{2}-1\right) d t \\
=\frac{2}{3}\left(\frac{t^{3}}{3}-t\right)+C=\frac{2}{3}\left[\frac{\left(1+x^{3}\right)^{3 / 2}}{3}-\left(1+x^{3}\right)^{\frac{1}{2}}\right]+C \\
=\frac{2}{9}\left(1+x^{3}\right)^{3 / 2}-\frac{2}{3}\left(1+x^{3}\right)^{1 / 2}+C
\end{array}$
Hence, the correct answer is option: 'D'.