<s>[INST] Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\, 

Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$

Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
 [/INST] Not an answer, strictly speaking, but an alternate method.
$$J=\int_{-\infty}^{\infty}\frac{x^2 dx}{(1+x^2)^2}=2\int_{0}^{\infty}\frac{x^2 dx}{(1+x^2)^2}$$
We may rewrite this as
$$\begin{align}
J&=2\int_0^\infty \frac{dx}{1+x^2}-2\int_0^\infty \frac{dx}{(1+x^2)^2}\\
&=\pi-2\int_0^\infty \frac{dx}{(1+x^2)^2}\ .
\end{align}$$ 
From here, we have that if
$$I_n=\int\frac{dx}{(ax^2+b)^n}\qquad a,b>0, n\in\Bbb N$$
then 
$$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}\qquad n>1\ ,$$
with the base case
$$I_1=\frac1{\sqrt{ab}}\arctan\sqrt{\frac{a}{b}}x+C\ .$$
Hence we have (with $a=b=n-1=1$)
$$\begin{align}
\int_0^\infty\frac{dx}{(x^2+1)^2}&=\frac{x}{2(x^2+1)}\bigg|_0^\infty+\frac{1}{2}\int_0^\infty\frac{dx}{x^2+1}\\
&=0+\frac12\cdot\frac\pi2\\
&=\frac\pi4
\end{align}$$
So 
$$\begin{align}
J&=\pi-2\cdot\frac\pi4\\
&=\frac\pi2
\end{align}$$
</s>