<s>[INST] 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.
 [/INST] $x \to 1/x$. We then have
$$I = \int_0^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_0^1\dfrac{\log(x)}{(1+x^2)^2}dx + \int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx$$
$$\int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{x^2\log(x)}{(1+x^2)^2}dx$$
Hence,
$$I = \int_0^1 \dfrac{1-x^2}{(1+x^2)^2}\log(x)dx = \int_0^1 (1-x^2)\left(\sum_{k=0}^{\infty} (k+1)(-1)^k x^{2k} \right)\log(x)dx$$
We now have
$$\int_0^1 x^{2n} \log(x) dx = -\dfrac1{(2n+1)^2}$$
Hence,
$$I = \sum_{k=0}^{\infty}(k+1)(-1)^k\left(-\dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2}\right)= -\dfrac{\pi}4$$
To get the last step, note that
$$(k+1)\left(\dfrac1{(2k+1)^2} - \dfrac1{(2k+3)^2}\right) = \dfrac14 \left(\dfrac1{2k+1} - \dfrac1{2k+3} + \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right)$$
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