Show that $\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50}$ Show that
$$\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50} $$
So I figured since it's an improper integral I should change the limits
$$\lim_{m_1\to-\infty}\int_{m_1}^{0}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}+ \lim_{m_2\to\infty}\int_{0}^{m_2}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}$$
I'm however not sure how to evaluate this. Any help would be great - thanks.       

Hint
Use partial fractions to write out the integral in the form:
$$\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E+F
   x}{x^2+2 x+2}$$
You'll find that $a=-6/25, c = 2/5$, etc. Now look at at what remains, and use the fact that:
$$\int \frac{1}{x^2+1}dx=\arctan x$$
And substitute $x\rightarrow x+1$ in the last integral to simplify it.
