$\int_{a}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)} \mathrm{d} x$,
解 我们有
\[\begin{aligned} \int \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x &=-\frac{1}{2} \int \ln x \mathrm{~d}\left(\frac{1}{1+x^{2}}\right) \\&=-\frac{\ln x}{2\left(1+x^{2}\right)}+\frac{1}{2} \int \frac{\mathrm{d} x}{x\left(1+x^{2}\right)} \\ &=-\frac{\ln x}{2\left(1+x^{2}\right)}+\frac{1}{2} \int\left(\frac{1}{x}-\frac{x}{1+x^{2}}\right) \mathrm{d} x \\&=-\frac{\ln x}{2\left(1+x^{2}\right)}+\frac{1}{4} \ln \frac{x^{2}}{1+x^{2}}+C . \\ \text { 由于 } & \lim _{\substack{c \rightarrow+0 \\ b \rightarrow+\infty}} \int_{a}^{b} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x \\=&\left.\lim _{\substack{-\rightarrow+0 \\ b \rightarrow+\infty}}\left[-\frac{\ln x}{2\left(1+x^{2}\right)}+\frac{1}{4} \ln \frac{x^{2}}{1+x^{2}}\right]\right|^{b} \\ =& \lim _{\substack{\epsilon \rightarrow+\infty \\ b \rightarrow+\infty}}\left[-\frac{\ln b}{2\left(1+b^{2}\right)}+\frac{\ln \varepsilon}{2\left(1+\varepsilon^{2}\right)}+\frac{1}{4} \ln \frac{b^{2}}{b^{2}+1}-\frac{1}{4} \ln \frac{\varepsilon^{2}}{\varepsilon^{2}+1}\right] \\ =& \lim _{\varepsilon \rightarrow+\infty}\left[-\frac{\varepsilon^{2}}{2\left(\varepsilon^{2}+1\right)} \ln \varepsilon+\frac{1}{4} \ln \left(\varepsilon^{2}+1\right)\right]=0, \\ \text { 所以, } & \int_{0}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x=0 . \end{aligned}\]