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Question Number 119960 by mnjuly1970 last updated on 28/Oct/20

Answered by mathmax by abdo last updated on 28/Oct/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{(1+\mathrm{x})\sqrt{\mathrm{x}}}\mathrm{dx}\Rightarrow \mathrm{A}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{(1+{\mathrm{t}}^2)\mathrm{t}}\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2}\mathrm{dt}{=}_{\mathrm{t}=\frac{1}{\mathrm{u}}}-2{\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\frac{1}{{\mathrm{u}}^2}\right)}{1+\frac{1}{{\mathrm{u}}^2}}\left(\frac{-\mathrm{du}}{{\mathrm{u}}^2}\right)$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-\arctan \left({\mathrm{u}}^2\right)}{1+{\mathrm{u}}^2}\mathrm{du}=\pi {\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{1+{\mathrm{u}}^2}-2{\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{u}}^2\right)}{1+{\mathrm{u}}^2}\mathrm{du}$$$$=\frac{{\pi }^2}{2}-\mathrm{A}\Rightarrow 2\mathrm{A}=\frac{{\pi }^2}{2}\Rightarrow \mathrm{A}=\frac{{\pi }^2}{4}\Rightarrow \sqrt{\mathrm{A}}=\frac{\pi }{2}$$

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