计算积分 ∫(0,1)ln(1+x)／(1+x^2) dx

elim

题：计算积分  \(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}dx\)

elim

题：计算积分\(\;\small\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}dx\)
解：作变量代换\(\,x=\tan\theta,\,\)则原积分
\(\qquad I={\small\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\int_0^{\large\frac{\pi}{4}}}\ln(1+\tan\theta)d\theta.\)
\(\qquad\overset{\varphi={\large\frac{\pi}{4}}-\theta}{=}{\small\displaystyle\int_0^{\large\frac{\pi}{4}}}\ln(1+\tan(\frac{\pi}{4}-\varphi))d\varphi\)
\(\because\quad(1+\tan\theta)(1+\tan(\frac{\pi}{4}-\theta))=2,\,\)取末二积分之平均得
\(\qquad I=\frac{1}{2}{\small\displaystyle\int_0^{\large\frac{\pi}{4}}}\ln 2\,d\theta = {\large\frac{\pi}{8}}\ln 2.\small\quad\square\)

luyuanhong

楼上 elim 的解答很好！已收藏。