<s>[INST] Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2}  \right) ^2 dx =  \lim_{ A \to \infty }  \int_{0}^{A} \left( \frac{x}{1+x^2}  \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2}  \right) ^2 dx $ but I don't have idea.
 [/INST] $$F(a)=\int-\frac{1}{1+ax^2}dx=-\frac{\arctan(\sqrt{a}x)}{\sqrt a}+c$$
$$\int_{-\infty}^{+\infty}-\frac{1}{1+ax^2}dx=-\frac{\pi}{\sqrt a}$$
so we take the derivaitve of $-\frac{\pi}{\sqrt a}$ with respect to $"a"$ then put a=1 so 
$$\int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{2}$$
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