Find: $$\int \dfrac{x^4+2x-1}{(x^2+1)^3} dx$$
Now I have attempted to use partial fractions to split the integral into three different fractions, but this has not helped me integrate this function, the main problem being the quadratic factor of $x^2$ in the denominator while lacking linear factors of $x$ in the numerator. What should I do?
Any help is highly appreciated.
$$\dfrac{x^4-1+2x}{(x^2+1)^3}=\dfrac{x^2-1}{(x^2+1)^2}+\dfrac{2x}{(x^2+1)^3}$$
For the second integral choose $x^2+1=y$
$$\dfrac{x^2-1}{(x^2+1)^2}=\dfrac{1-\dfrac1{x^2}}{\left(x+\dfrac1x\right)^2}$$
$\displaystyle\int\left(1-\dfrac1{x^2}\right)dx=?$