Evaluate the integral $PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$ Using the Cauchy Integral Principal to evaluate: 
$$PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$$
I know this integral has a pole at $x=i$, $x = -1-i$ and $x = -1+i$. Can someone please show me how to use Cauchy Principal value to evaluate this integral ?

Let $C_R$ denote the closed semi-circle of radius $R$ centered at $z=0$ in the lower-half plane oriented in anti-clockwise fashion.
$$
\int_{C_R}{dz\over(z^2+1)(z^2+2z+2)}=\\\int_{\pi}^{2\pi}{iRe^{it}dt\over(R^2e^{2it}+1)(R^2e^{2it}+2Re^{it}+2)}+\int_{R}^{-R}{dx\over(x^2+1)(x^2+2x+1)}
$$
Now $|R^2e^{2it}+1|=R^2|e^{2it}+R^{-2}|>{R^2\over2}$ for large enough $R$. 
Simliarly,
$|R^2e^{2it}+2Re^{it}+2|>{R^2\over2}$ for large enough $R$.
So for large enough $R$, the absolute value of the integrand in the first integral is less than ${4\over R^3}$.
$$
\therefore\lim_{R\to\infty}\int_{C_R}{dz\over(z^2+1)(z^2+2z+2)}=\int_{\infty}^{-\infty}{dx\over(x^2+1)(x^2+2x+1)}
$$ 
As Winther notes Cauchy principal value is redundant here. Using residue theorem we have
$$
\int_{-\infty}^{\infty}{dx\over(x^2+1)(x^2+2x+1)}=-2\pi i\left[\text{Res}_{-1-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)+\text{Res}_{-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)\right]=-2\pi i{1\over(z^2+1)(z+1-i)}\Big{|}_{z=-1-i}-2\pi i{1\over(z-i)(z^2+2z+2)}\Big{|}_{z=-i}\\
={2\pi\over5}
$$
