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Question Number 90279 by mathmax by abdo last updated on 22/Apr/20

$$calculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x^2\right)}{x^2+4}dx$$

Commented by mathmax by abdo last updated on 23/Apr/20

$$A={\int }_0^{\mathrm{\infty }}\frac{arctan\left(x^2\right)}{x^2+4}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(x^2\right)}{x^2+4}dxlet$$$$\phi \left(z\right)=\frac{arctan\left(z^2\right)}{z^2+4}\Rightarrow \phi \left(z\right)=\frac{arctan\left(z^2\right)}{(z-2i)(z+2i)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2i)=2i\pi \frac{\mid arctan(-4)\mid }{4i}$$$$=\frac{\pi }{2}arctan\left(4\right)\Rightarrow A=\frac{\pi }{4}arctan\left(4\right)(lookthat{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x^2\right)}{x^2+4}dx>0)$$

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