Question

If $I =\int\frac{x^{5}}{\sqrt{1+x^{3}}}dx$ , 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. $log\left|\sqrt{x}-\sqrt{1+x^{3}}\right|+C$
• D. $\frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+C$

Answer 1

Answer: (D)

$I =\int\frac{x^{5}}{\sqrt{1+x^{3}}}dx=\int \frac{x^{3}.x^{2}}{\sqrt{1^{.}+x^{3}}}dx$
$Let 1 + x^{3} = t^{2}, so that 3x^{2} dx = 2t dt$
$\Rightarrow x^{2} dx = \frac{2}{3}td t$
$\therefore I =\int\frac{\left(t^{2}-1\right)\frac{2}{3}t dt}{t}=\frac{2}{3}\int t^{2} -1)dt$
$=\frac{2}{3}\left(\frac{t^{3}}{3}-t\right)+C =\frac{2}{3}\left[\frac{\left(1+x^{3}\right)^{3/2}}{3}-\left(1+x^{3}\right)^{\frac{1}{2}}\right]+C$
$=\frac{2}{9}\left(1+x^{3}\right)^{3/2}-\frac{2}{3}\left(1+x^{3}\right)^{1/2}+C$