<s>[INST] Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$

The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.

Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
 [/INST] Consider the function $\displaystyle f(x) = \frac{1-x^2}{(1+x^2)^2} $. The Maclaurin series of $f$ is given by:
$$\displaystyle \sum_{k \ge 0}  [k(-1)^k x^{2k} - k(-1)^k x^{2k-2}]$$
The integral is equal to $\displaystyle \int_0^1 f(x)\log{x}\,\mathrm{d}x $. Using the series expansion you should get
$$\displaystyle I = \sum_{k \ge 0} \bigg(\frac{k(-1)^k}{(2k-1)^2} + \frac{k(-1)^{k+1}}{(2k+1)^2}\bigg) = \sum_{k \ge 0} \frac{(-1)^{k+1}}{(2k+1)} = -\frac{\pi}{4}.$$
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