<s>[INST] Find $\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx$ 
Evaluate $$
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx
$$

$$
\int\frac{x}{(1+x^2)^2}dx=\frac{-1}{2(1+x^2)}\\
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx=
\bigg[\log x\frac{-1}{2(1+x^2)}\bigg]_0^\infty-\int_0^\infty\frac{1}{x}.\frac{-1}{2(1+x^2)}dx\\
=0+\int_0^\infty\frac{1}{2x(1+x^2)}dx=\frac{1}{2}\int\bigg[\frac{1}{x}-\frac{x}{1+x^2}\bigg]dx\\
=\frac{1}{2}[\log x]^\infty_0-\frac{1}{4}[\log|1+x^2|]_0^\infty=\frac{1}{4}\log\frac{x^2}{1+x^2}\\
=\bigg[\frac{1}{4}\log\frac{1}{\dfrac{1}{x^2}+1}\bigg]_0^\infty
$$
Is there a better substitution that I can make to evaluate thi definite integral ?
Note: The solution given in my reference is $0$
Similar question is asked @Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ but that is about the convergence of the given integral, here I am looking for a better and obvious substitution that will make the above definite integral easier to solve. But, hint about the solution is found there, Thanks @lab bhattacharjee
 [/INST] Note that, with $x=\frac1t$,
$$\int_1^\infty\frac{x\log x}{(1+x^2)^2}dx
=- \int_0^1\frac{t\log t}{(1+t^2)^2}dt$$
Thus, the integral is evaluated to zero.
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