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Question Number 30184 by abdo imad last updated on 17/Feb/18

$$find{\int }_{\frac{1}{2}}^2(1+\frac{1}{x^2})arctanxdx.(arctan={tan}^{-1}).$$

Commented by abdo imad last updated on 20/Feb/18

$$thech.x=\frac{1}{t}giveI={\int }_{\frac{1}{2}}^2(1+t^2)arctan\left(\frac{1}{t}\right)\frac{dt}{t^2}$$$$={\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})(\frac{\pi }{2}-arctant)dt$$$$=\frac{\pi }{2}{\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})dt-{\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})arctantdt\Rightarrow 2I=\frac{\pi }{2}{\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})dt$$$$2I=\frac{\pi }{2}\frac{3}{2}+\frac{\pi }{2}{[-\frac{1}{t}]}_{\frac{1}{2}}^2=\frac{3\pi }{4}+\frac{\pi }{2}(2-\frac{1}{2})=\frac{3\pi }{4}+\frac{3\pi }{4}=\frac{3\pi }{2}\Rightarrow$$$$I=\frac{3\pi }{4}.$$

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