Question

If the integral $I=\int \frac{x^5}{\sqrt{1+x^3}}dx$ $=K\sqrt{x^3+1}\left(x^3-2\right)+C,$ (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 $1+x^3=t^2$
$\Rightarrow 3x^{2}dx=2tdt$
$\Rightarrow I=\int \frac{\left(t^2-1\right)2tdt}{3t}$
$\Rightarrow I=\frac{2}{3}\int \left(t^2-1\right)dt$
$=\frac{2t^3}{9}-\frac{2}{3}t+C$
$=\frac{2}{9}t\left(t^2-3\right)+C$
$=\frac{2}{9}\sqrt{1+x^3}\left(x^3-2\right)+C$
$\therefore K=\frac{2}{9}\Rightarrow 9K=2$