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Question Number 137770 by peter frank last updated on 06/Apr/21

Answered by Ñï= last updated on 06/Apr/21

$$I={\int }_0^1\frac{ln(1+x)}{1+x^2}dx={\int }_0^{1}da{\int }_0^1\frac{x}{(1+ax)(1+x^2)}dx$$$$={\int }_0^1\frac{1}{1+a^2}da{\int }_0^1(\frac{-a}{1+ax}+\frac{x+a}{1+x^2})dx$$$$={\int }_0^1\frac{1}{1+a^2}da{(-ln(1+ax)+\frac{1}{2}ln(1+x^2)+a{\tan }^{-1}x)}_0^1$$$$={\int }_0^1\frac{1}{1+a^2}(-ln(1+a)+\frac{1}{2}ln2+\frac{\pi }{4}a)da$$$$=\frac{\pi }{4}ln2-I$$$$\Rightarrow I=\frac{\pi }{8}ln2$$

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