# How to use partial fractions with a cubic factor on bottom?

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.

• Your title says that we must use partial fractions, but the main text doesn't. Jul 30 '16 at 7:34
• You don't have to, but it is the easiest way to do so. Jul 30 '16 at 7:43

$$\dfrac{x^4-1+2x}{(x^2+1)^3}=\dfrac{x^2-1}{(x^2+1)^2}+\dfrac{2x}{(x^2+1)^3}$$
For the second integral choose $x^2+1=y$
$$\dfrac{x^2-1}{(x^2+1)^2}=\dfrac{1-\dfrac1{x^2}}{\left(x+\dfrac1x\right)^2}$$
$\displaystyle\int\left(1-\dfrac1{x^2}\right)dx=?$
• @JohnSmith, Why? $$x^4+2x-1=(x^2-1)(x^2+1)+2x$$ Jul 30 '16 at 7:41