# Integrate the functions$\frac{3x^2}{x^6+1}$

Toolbox:
• $(i)\;\int \frac{dx}{x^2+1}=\tan^{-1}x+c.$
Given $I=\int\frac{3x^2}{x^6+1}dx.$

Let $x^6=(x^3)^2.$

$\;\;\;=\int\frac{3x^2}{(x^3)^2+1}.$

Let $x^3=t.$

On differentiating we get,

$3x^2dx=dt.$

Substituting this we get,

$I=\int\frac{dt}{(t^3)^2+1}.$

On integrating we get,

$\;\;\;=\tan^{-1}t+c.$

substituting for t we get,

$\frac{3x^2}{x^6+1}=\tan^{-1}(x^3)+c.$
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