Question

Prove that
$\int \frac{x^2}{x^6+1}dx$

Answer 1

We have,

$\begin{array}{c}\int \frac{x^2}{x^6+1}dx\\ put{x}^3=t3x^{2}dx=dt\\ =\int \frac{\frac{dt}{3}}{t^2+1}\\ =\frac{1}{3}{\tan }^{-1}t+c\\ =\frac{1}{3}{\tan }^{-1}{x}^3+c\end{array}$

Hence, this is the answer.