微积分(A)每日一题[38]

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

\[I=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx \\ \xlongequal{x=\tan t} \int_0^{\frac{\pi}{4}} \frac{\ln(1+\tan t)}{\sec^2t} \frac{dt}{\cos^2 t} \\ =\int_0^\frac{\pi}{4} \ln(1+\tan t) dt \\\xlongequal{\tan (\frac{\pi}{4}-t)=\frac{1-\tan t}{1+\tan t}} \int_0^{\frac{\pi}{4}}\ln(1+\tan(\frac{\pi}{4}-t))dt \\ =\int_0^{\frac{\pi}{4}}\ln \frac{2}{1+\tan t} dt \\ =\int_0^{\frac{\pi}{4}}\ln2dt-\int_0^{\frac{\pi}{4}}\ln(1+\tan t)dt \\ =\frac{\pi\ln 2}{4}-I \\ \Rightarrow I=\frac{\pi\ln 2}{8} \]

标签：38,frac,ln,微积分,int,dt,tan,pi,每日
来源： https://www.cnblogs.com/nekko/p/15720862.html