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Question Number 159474 by mnjuly1970 last updated on 17/Nov/21

$$$$$$niceintegral.$$$$provethat$$$${\underset{0}{\int }}^{\mathrm{\infty }}\frac{{tan}^{-1}\left(2x\right)+{tan}^{-1}\left(\frac{x}{2}\right)}{1+x^2}dx=\frac{{\pi }^2}{4}$$$$$$$$$$$$$$

Answered by qaz last updated on 18/Nov/21

$${\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(2\mathrm{x}\right)+{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\frac{2}{\mathrm{x}}+{\tan }^{-1}\frac{\mathrm{x}}{2}}{1+{\mathrm{x}}^2}\mathrm{dx}.....2\mathrm{x}\to \frac{2}{\mathrm{x}}$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\frac{{\pi }^2}{4}$$

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