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Question Number 163536 by HongKing last updated on 07/Jan/22

$$\mathrm{Prove}\mathrm{that}:$$$$\underset{9}{\stackrel{1}{\int }}\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}=\frac{\pi \cdot \ln \left(2\right)}{8}$$

Answered by Ar Brandon last updated on 07/Jan/22

$$I={\int }_0^1\frac{\ln (1+x)}{1+x^2}dx,x=\tan \vartheta$$$$={\int }_0^{\frac{\pi }{4}}\ln (1+\tan \vartheta )d\vartheta$$$$={\int }_0^{\frac{\pi }{4}}\ln (\cos \vartheta +\sin \vartheta )d\vartheta -{\int }_0^{\frac{\pi }{4}}\ln \left(\cos \vartheta \right)d\vartheta$$$$={\int }_0^{\frac{\pi }{4}}\ln \left(\sqrt{2}\cos (x-\frac{\pi }{4})\right)dx-{\int }_0^{\frac{\pi }{4}}\ln \left(\cos \vartheta \right)d\vartheta$$$$=\frac{\pi \ln 2}{8}+{\int }_0^{\frac{\pi }{4}}\ln \left(\cos (x-\frac{\pi }{4})\right)-{\int }_0^{\frac{\pi }{4}}\ln \left(\cos \vartheta \right)d\vartheta$$$$=\frac{\pi \ln 2}{8}+{\int }_0^{\frac{\pi }{4}}\ln \left(\cos \vartheta \right)d\vartheta -{\int }_0^{\frac{\pi }{4}}\ln \left(\cos \vartheta \right)d\vartheta =\frac{\pi \ln 2}{8}$$

Commented by HongKing last updated on 07/Jan/22

$$\mathrm{perfect}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by peter frank last updated on 11/Jan/22

$$\mathrm{great}$$

Answered by smallEinstein last updated on 07/Jan/22

Commented by GalaxyBills last updated on 07/Jan/22

$$MyBossthat$$

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