Question

If the integral $I=\displaystyle \int \frac{x^{5}}{\sqrt{1 + x^{3}}}dx$ $=K\sqrt{x^{3} + 1}\left(x^{3} - 2\right)+C,$ (where, $C$ is the constant of integration), then the value of $9K$ is equal to

• A. $4$
• B. $2$
• C. $6$
• D. $10$

Answer 1

Answer: (B)

Let $1+x^{3}=t^{2}$
$\Rightarrow 3x^{2}dx=2tdt$
$\Rightarrow I=\displaystyle \int \frac{\left(t^{2} - 1\right) 2 t d t}{3 t}$
$\Rightarrow I=\frac{2}{3}\displaystyle \int \left(t^{2} - 1\right)dt$
$=\frac{2 t^{3}}{9}-\frac{2}{3}t+C$
$=\frac{2}{9}t\left(t^{2} - 3\right)+C$
$=\frac{2}{9}\sqrt{1 + x^{3}}\left(x^{3} - 2\right)+C$
$\therefore K=\frac{2}{9}\Rightarrow 9K=2$