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Question Number 184626 by SANOGO last updated on 09/Jan/23

$${\int }_0^{+oo}\frac{artan\left(2x\right)-arctan\left(x\right)}{x}dx$$

Answered by qaz last updated on 10/Jan/23

$${\int }_0^{\mathrm{\infty }}\frac{\arctan 2x-\arctan x}{x}dx$$$$={\int }_0^{\mathrm{\infty }}{\int }_1^2\frac{1}{1+a^2{x}^2}dadx={\int }_1^{2}da{\int }_0^{\mathrm{\infty }}\frac{dx}{1+a^2{x}^2}$$$$={\int }_1^2\frac{\pi }{2a}da=\frac{\pi }{2}ln2$$

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