Question

Find :
$\int \frac{{3x}^2}{x^6+1}dx$

Answer 1

$\int \frac{3x^{2}dx}{x^6+1}$

$=\int \frac{3x^{2}dx}{{\left(x^3\right)}^2+1}$

$t=x^3\Rightarrow dt=3x^{2}dx$

$=\int \frac{dt}{t^2+1}$

$={\tan }^{-1}t+c$ where $c$ is the constant of integration.

$={\tan }^{-1}\left(x^3\right)+c$ where $t=x^3$