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Question Number 134182 by Algoritm last updated on 28/Feb/21

Answered by Ñï= last updated on 01/Mar/21

$${\int }_0^1{\left({\tan }^{-1}\frac{1-x}{1+x}\right)}^3\frac{(1+x)}{(1-x)(1+x^2)}dx$$$$={\int }_0^1{\left({\tan }^{-1}\frac{1-x}{1+x}\right)}^3\frac{1+x}{1-x}d{\tan }^{-1}x$$$$={\int }_0^1{\left({\tan }^{-1}\frac{1-x}{1+x}\right)}^3\frac{1+x}{1-x}d({\tan }^{-1}1-{\tan }^{-1}\frac{1-x}{1+x})$$$$=-{\int }_0^1{\left({\tan }^{-1}\frac{1-x}{1+x}\right)}^3\frac{1+x}{1-x}d\left({\tan }^{-1}\frac{1-x}{1+x}\right)$$$$=-\frac{1}{4}{\int }_0^1\frac{1+x}{1-x}d{\left({\tan }^{-1}\frac{1-x}{1+x}\right)}^4$$$$....continue.....$$

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