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Question Number 86646 by Ar Brandon last updated on 30/Mar/20

$${\int }_0^1\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$$$$$$$$$$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$I={\int }_0^1\frac{ln(1+x)}{1+x^2}dxcha7gementx=tantgive$$$$I={\int }_0^{\frac{\pi }{4}}\frac{ln(1+tan\left(t\right))}{1+{tan}^{2}t}(1+{tan}^{2}t)dt$$$$={\int }_0^{\frac{\pi }{4}}ln(1+tant)dt$$$${=}_{t=\frac{\pi }{4}-u}{\int }_0^{\frac{\pi }{4}}ln(1+tan(\frac{\pi }{4}-u))du$$$$={\int }_0^{\frac{\pi }{4}}ln(1+\frac{1-tanu}{1+tanu})du={\int }_0^{\frac{\pi }{4}}ln\left(\frac{2}{1+tanu}\right)du$$$$=\frac{\pi }{4}ln\left(2\right)-{\int }_0^{\frac{\pi }{4}}ln(1+tanu)du\Rightarrow$$$$2I=\frac{\pi }{4}ln\left(2\right)\Rightarrow I=\frac{\pi }{8}ln\left(2\right)$$

Commented by Ar Brandon last updated on 30/Mar/20

$$cool!!$$$$$$

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