[JamesCoe#18]

ข้อความเดิมเขียนโดยคุณ Mastermander

$\displaystyle{\int_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{x^2 + 1}}dx} }$
Let $x=\tan u$ , then integral becomes
\[
\int_0^{\frac{\pi }{4}} {\ln \left( {\tan u + 1} \right)\,du}
\]
let \[
u = \frac{\pi }{4} - v
\]
thus
\[
\int_0^{\frac{\pi }{4}} {\ln \left( {\tan u + 1} \right)\,du} = \int_0^{\frac{\pi }{4}} {\ln \left( {\tan \left( {\frac{\pi }{4} - v} \right) + 1} \right)\,dv}
\]
\[
\tan \left( {A - B} \right) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}
\]
\[
\int_0^{\frac{\pi }{4}} {\ln \left( {\tan \left( {\frac{\pi }{4} - v} \right) + 1} \right)\,dv} = \int_0^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \tan v}}{{1 + \tan v}} + 1} \right)\,dv} = \int_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan v}}} \right)\,dv}
\]
\[
\int_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan v}}} \right)\,dv} = \int_0^{\frac{\pi }{4}} {\ln 2 - \ln \left( {1 + \tan v} \right)\,dv}
\]
\[
\int_0^{\frac{\pi }{4}} {\ln \left( {\tan u + 1} \right)\,du} = \frac{\pi }{4}\ln 2 - \int_0^{\frac{\pi }{4}} {\ln \left( {1 + \tan v} \right)\,dv}
\]
As a result \[
\int_0^{\frac{\pi }{4}} {\ln \left( {\tan u + 1} \right)\,du} = \frac{\pi }{8}\ln 2
\]


แถมให้อีกข้อ
Given $a>0$. Evaluate $\displaystyle{ \int_{\frac{1}{a}}^a \frac{\tan^{-1}x}{x} \,\,d x }$

ปล. โจทย์ที่ผมโพสมีเฉลยอยู่แล้วในที่นี่ถ้าอยากรู้คำตอบเร็วก็ลองค้นหาดูก็ได้ครับ

ยากไปไหมคับขอแค่ธรรมดาๆก่อนนะคับ