# 问题及解答

## 计算定积分 $\displaystyle\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x$.

Posted by haifeng on 2023-03-25 14:10:45 last update 2023-03-25 14:11:06 | Edit | Answers (2)

1

Posted by haifeng on 2023-03-25 14:17:34

$\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x=\int_0^{1}\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x+\int_1^{\infty}\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x$

$\begin{split} \int_1^{\infty}\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x&=\int_1^0 \frac{\frac{1}{t}\ln\frac{1}{t}}{(1+\frac{1}{t^2})^2}\mathrm{d}\frac{1}{t}\\ &=\int_1^0 \frac{-\ln t}{t(1+\frac{1}{t^2})^2}\cdot\frac{-1}{t^2}\mathrm{d}t\\ &=-\int_0^1\frac{\ln t}{t^3\cdot\frac{(t^2+1)^2}{t^4}}\mathrm{d}t\\ &=-\int_0^1\frac{t\ln t}{(1+t^2)^2}\mathrm{d}t \end{split}$

2

Posted by haifeng on 2023-03-25 14:25:48

(法二) 也可以先求被积函数的原函数.

$\begin{split} \int\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x&=-\frac{1}{2}\int\ln x\mathrm{d}\frac{1}{1+x^2}\\ &=-\frac{1}{2}\biggl[\frac{\ln x}{1+x^2}-\int\frac{1}{1+x^2}\mathrm{d}\ln x\biggr]\\ &=-\frac{\ln x}{2(1+x^2)}+\frac{1}{2}\cdot\int\frac{1}{1+x^2}\cdot\frac{1}{x}\mathrm{d}x, \end{split}$

$\begin{split} \int\frac{1}{1+x^2}\cdot\frac{1}{x}\mathrm{d}x&=\int(\frac{1}{x}-\frac{x}{1+x^2})\mathrm{d}x\\ &=\ln x-\int\frac{x}{1+x^2}\mathrm{d}x\\ &=\ln x-\frac{1}{2}\int\frac{1}{1+x^2}\mathrm{d}x^2\\ &=\ln x-\frac{1}{2}\ln(1+x^2)+C \end{split}$

$\begin{split} \int\frac{x\ln x}{(1+x^2)^2}\mathrm{d}x&=-\frac{\ln x}{2(1+x^2)}+\frac{1}{2}\ln x-\frac{1}{4}\ln(1+x^2)+C\\ &=\frac{x^2\ln x}{2(1+x^2)}-\frac{1}{4}\ln(1+x^2)+C \end{split}$