微积分(A)每日一题[40]

计算：\(\int_0^{+\infty}\frac{\ln x}{(1+x)^2}dx\)

\[\begin{aligned} I&=\int_0^{+\infty} \frac{\ln x}{(1+x)^3}dx \\ &=-\int_0^{+\infty} \frac{\ln \frac{1}{x}}{(1+\frac{1}{x})^3}\left(-\frac{1}{x^2}\right)dx \\ &=-\int_0^{+\infty} \frac{\ln x}{(1+x)^3}xdx \\ \Rightarrow&\int_0^{+\infty} \frac{\ln x}{(1+x)^2}=I-I=0 \end{aligned} \]