$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$
$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$
$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$
$= xlnx.ln(x^{2}+1)^{2}-c+H$
$H=\int\limits_{}^{}2ln(x^{2}+1).(lnx+1)dx$
$= \int\limits_{}^{}[2ln(x^{2}+1)lnx]dx+\int\limits_{}^{}2ln(x^{2}+1)dx$
$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$
$=2(x+x^{2}+1)ln(x+x^{2}+1)-(x+x^{2}+1)+2(x^{2}+1)ln(x^{2}+1)-(x^{2}+1)+c$
" Công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ "
$(1)=..................... $