How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$
I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was wondering if there is a faster technique to resolve this.
$$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx\\ =\int\left(+\frac{2x+2}{2(x^2+2x+2)}+\frac{2x+2}{2(x^2+2x-2)}-\frac1{\frac{(2x+2)^2}4+1}\right)\mathrm dx\\ =\frac12\ln(x^2+2x+2)+\frac12\ln(x^2+2x-2)-\arctan\left(\frac{2x+2}2\right)+\mathbb C$$
HINT:
Write $$\frac{2x^3 + 5x^2 +2x +2)}{(x^2 +2x + 2)(x^2 + 2x - 2)}=\frac{Ax+B}{x^2+2x+2}+\frac{Cx+D}{x^2+2x-2}$$
For the ease of calculation, we can write
$$\frac{Ax+B}{x^2+2x+2}=\frac A2\cdot\frac{\left(\dfrac{d(x^2+2x+2)}{dx}\right)}{x^2+2x+2}+\frac C{(x+1)^2+1^2}.$$
