Question

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

Answer 1

Consider the given integral.

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

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

Let $t=x^3$

$dt=3x^{2}dx$

Therefore,

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

$I={\tan }^{-1}t+C$

On putting the value of $t$, we get

$I={\tan }^{-1}\left(x^3\right)+C$

Hence, this is the answer.