<s>[INST] Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$ $$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx
$$
Help me please. I don't know any ways of solution. Thank you.
 [/INST] Integrating by Parts,
$$I=\int \ln x\cdot\frac x{(1+x^2)^2}dx$$
$$=\ln x\int\frac x{(1+x^2)^2}dx-\int\left(\frac{d(\ln x)}{dx}\cdot \frac x{(1+x^2)^2}dx\right)dx $$
For $\displaystyle J=\int\frac x{(1+x^2)^2}dx$ set $1+x^2=u\implies2x\ dx=du$
$\displaystyle J=\int\frac{du}{2u^2}=-\frac1{2u}=-\frac1{2(1+x^2)}$
$\displaystyle\implies I=\ln x\left(-\frac1{2(1+x^2)}\right)+\int\frac1{2x(1+x^2)}\ dx$
Setting $x^2=v,$
$\displaystyle K=\int\frac1{2x(1+x^2)}\ dx=\int\frac{2x}{4x^2(1+x^2)}\ dx$
$\displaystyle4K=\int\frac{dv}{v(1+v)}=\int\frac{v+1-v}{v(1+v)}dv=\int\frac{dv}v -\int\frac{dv}{v+1}=\cdots$
</s>