Evaluate $\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$ I am supposed to evaluate
$$\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$$
I started using partial fractions
$$3x^5+13x^4+32x^3+8x^2-40x-75=x\left(x^2+3x+5^2\right)^2A+\left(x^2+3x+5^2\right)^2B+x\left(x^2+3x+5^2\right)\left(Cx+D\right)+x^2\left(Ex+F\right)$$
I managed to get to
$$\:\int \:\left(\frac{2}{x}-\frac{3}{x^2}+\frac{x+1}{x^2+3x+5}+\frac{4x}{\left(x^2+3x+5\right)^2}\right)\:dx$$
Am i on the right track? is there an easier way to simplify the original integral?

An alternative method to treat multiple roots is the following (see Application to symbolic integration):
$$
\frac{3 x^5 + 13 x^4 + 32 x^3 + 8 x^2 - 40 x - 75}{x^2(x^2+3x+5)^2}=\frac{a}{x}+\frac{bx+c}{x^2+3x+5}+\frac{d}{dx}\left[\frac{e}{x}+\frac{fx+g}{x^2+3x+5}\right]
$$
from which we get the linear system
$$
\left\{
\begin{align}
&a+b=3,\\
&6 a+3 b+c-e-f=13,\\
&19 a+5 b+3 c-6 e-2 g=32,\\
&30 a+5 c-19 e+5 f-3 g=8,\\
&25 a-30 e=-40,\\
-&25 e=-75
\end{align}
\right.
$$
Finally, we obtain
$$
\int\left(\frac{2}{x}+\frac{x-\frac{1}{11}}{x^2+3 x+5}\right)dx+\frac{3}{x}+\frac{-\frac{12}{11}x-\frac{40}{11}}{x^2+3 x+5}
$$
The subsequent integration is pretty easy.
