Đặt \(\left\{\begin{array}{l} {u=\ln x} \\ {{\rm d}v=\frac{2x}{\left(x^{2} +1\right)^{2} } {\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=\frac{1}{x} {\rm d}x} \\ {v=-\frac{1}{x^{2} +1} } \end{array}\right. .\)
Khi đó:\(\int \frac{2x\ln x}{\left(x^{2} +1\right)^{2} } {\rm d}x =-\frac{\ln x}{x^{2} +1} +\int \frac{1}{x\left(x^{2} +1\right)} {\rm d}x =-\frac{\ln x}{x^{2} +1} +\int \frac{\left(x^{2} +1\right)-x^{2} }{x\left(x^{2} +1\right)} {\rm d}x .
\)
\(=-\frac{\ln x}{x^{2} +1} +\int \left(\frac{1}{x} -\frac{x}{x^{2} +1} \right) {\rm d}x.\)
\(=-\frac{\ln x}{x^{2} +1} +\ln x-\frac{1}{2} \ln \left(x^{2} +1\right)+C.\)