当前位置：网站首页>∫(0→1) ln(1+x) / (x ² + 1) dx

∫(0→1) ln(1+x) / (x ² + 1) dx

2022-06-27 21:43:00 【Super big surpass】

$\int_{0}^{1}\frac{\ln \left ( 1+x \right )}{1+x^{2}} dx=\int_{0}^{1}\ln \left ( 1+x \right ) d\left ( \arctan x \right )$

Make x=tan u, Then the original formula

$I=\int_{0}^{\frac{\pi }{4}}\ln \left ( 1+\tan u \right ) du$

$=\int_{0}^{\frac{\pi }{4}}\ln \left ( 1+\tan \left (\frac{\pi }{4} -u \right ) \right ) du$ 【 Interval reproduction formula 】

$=\int_{0}^{\frac{\pi }{4}}\ln \left ( 1+ \frac{1-\tan u}{1+\tan u} \right ) du=\int_{0}^{\frac{\pi }{4}}\ln \left ( \frac{2}{1+\tan u} \right ) du$

$=\int_{0}^{\frac{\pi }{4}}\ln 2 - \ln \left ( 1+\tan u \right ) du$

namely

$I=\frac{1}{2}\int_{0}^{\frac{\pi }{4}}\ln 2du$

$=\frac{\pi\ln 2 }{8}$

版权声明
本文为[Super big surpass]所创，转载请带上原文链接，感谢
https://yzsam.com/2022/178/202206271924398368.html