Question

Integrate the following functions.
$\int \frac{3x^2}{x^6+1}dx.$

Answer 1

$\int \frac{3x^2}{x^6+1}dx=\int \frac{3x^2}{(x^3{)}^2+1}dx$
Let $x^3=t\Rightarrow 3x^2=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{3x^2}$
$\therefore \int \frac{3x^2}{(x^3{)}^2+1}dx=\int \frac{3x^2}{t^2+1}.\frac{dt}{3x^2}=\int \frac{dt}{t^2+1}=\frac{1}{1}ta{n}^{-1}\left(\frac{t}{1}\right)+C=ta{n}^{-1}\left(x^3\right)+C$