Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.

Once you do the variable change in Ron Gordon's answer there's a trick that finishes it off pretty quickly. The integral is reduced to
$$\int_0^1 {1 - x^2 \over (1 + x^2)^2}\log(x)\,dx$$
$$= \int_0^1 {{1\over x^2} - 1 \over ({1 \over x}  + x)^2}\log(x)\,dx$$
Note that ${\displaystyle {{1\over x^2} - 1 \over ({1 \over x}  + x)^2}}$ is the derivative of ${\displaystyle{1 \over {1 \over x } + x}}$. So it's natural to integrate by parts in the above, and get
$$-\int_0^1{1 \over {1 \over x} + x} {1 \over x}\,dx$$
(The boundary terms here are both zero). This is the same as
$$-\int_0^1 {1 \over 1 + x^2}\,dx$$
$$= -{\pi \over 4}$$
