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 ?
 A: There is no need of using the Cauchy Principal Value here, because the function 
$$
f: \mathbb{C} \to \mathbb{C}, f(z)=\frac{1}{(z^2+1)(z^2+2z+2)}
$$ is defined everywhere on $\mathbb{R}$ (its denominator is nowhere vanishing on $\mathbb{R}$). You should use the Residue Theorem instead. 
Notice first that $f$ has four simple poles:
$$
-i,\, i,\, -1-i,\, -1+i.
$$
Given $R>\sqrt2$, we denote by $\Delta(R)$ the bounded region of $\mathbb{C}$ whose boundary consists of the segment $[-R,R]$, and the upper half circle centered at the origin with radius $R$.
Thanks to the Residue Theorem, we have:
\begin{eqnarray}
\int_{\Delta(R)}f(z)\,dz&=&2\pi i[\mathrm{Res}(f,i)+\mathrm{Res}(f,-1+i)]=2\pi i\left[\frac{1}{2i(i^2+2i+2)}+\frac{1}{[(-1+i)^2+1][2(-1+i+1)]}\right]\\
&=&2\pi \left[\frac{1}{2(1+2i)}+\frac{1}{2(1-2i)}\right]=\frac{2\pi}{5}.
\end{eqnarray}
Now, notice that
$$
\int_{\Delta(R)}f(z)\,dz=\int_{-R}^Rf(x)\,dx+i\int_0^\pi Re^{it}f(Re^{it})\,dt,
$$
and
$$
\lim_{R\to\infty}\left|i\int_0^\pi Re^{it}f(Re^{it})\,dt\right|\le \lim_{R\to\infty}\frac{\pi R}{(R^2-1)(R^2-2R-2)}=0,
$$
and therefore
$$
\int_{-\infty}^\infty\frac{1}{(x^2+1)(x^2+2x+2)}\,dx=\lim_{R\to\infty}\int_{-R}^Rf(x)\,dx=\lim_{R\to\infty}\int_{\Delta(R)}f(z)\,dz=\frac{2\pi}{5}.
$$
A: 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
$$
\int_{-\infty}^{\infty}{dx\over(x^2+1)(x^2+2x+1)}=-2\pi i\left[\text{Res}_{-1-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)+\text{Res}_{-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)\right]=-2\pi i{1\over(z^2+1)(z+1-i)}\Big{|}_{z=-1-i}-2\pi i{1\over(z-i)(z^2+2z+2)}\Big{|}_{z=-i}\\
={2\pi\over5}
$$
