Question

If $\int \frac{x^{5}dx}{\sqrt{1+x^3}}=\frac{2}{9}\sqrt{a+x^3}(x^3-b)+C$, then the value of $2a+3b$ is equal to

Answer 1

Given:

$I=\int \frac{x^5}{\sqrt{1+x^3}}dx$

Substitute $1+x^3=t^2$$\Rightarrow 3x^{2}dx=2tdt$

$I=\int \frac{t^2-1}{t}\left(\frac{2}{3}tdt\right)=\frac{2}{3}\int (t^2-1)dt$
$=\frac{2}{3}(\frac{t^3}{3}-t)+C=\frac{2}{9}t(t^2-3)+C$
$=\frac{2}{9}\left(\sqrt{1+x^3}\right)(x^3-2)+C$
$\therefore a=1,b=2$
$\therefore 2a+3b=8$

Hence, answer is $8$.