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Question Number 27805 by abdo imad last updated on 15/Jan/18

$$find{\int }_1^{\propto }\frac{arctan\left(\alpha x\right)}{x^2}.$$

Commented by abdo imad last updated on 16/Jan/18

$$letputI={\int }_1^{\propto }\frac{arctan\left(\alpha x\right)}{x^2}dxintegrateperparts$$$$I={[-\frac{1}{x}arctan\left(\alpha x\right)]}_1^{+\propto }-{\int }_1^{+\propto }-\frac{1}{x}\frac{\alpha }{1+{\alpha }^2{x}^2}dx$$$$I=artan\left(\alpha \right)+\alpha {\int }_1^{+\propto }\frac{dx}{x(1+{\alpha }^2{x}^2)}weusethech.x=\frac{1}{\alpha }t$$$${\int }_1^{+\propto }\frac{dx}{x(1+{\alpha }^2{x}^2)}={\int }_{\alpha }^{+\propto }\frac{\frac{1}{\alpha }dt}{\frac{1}{\alpha }t(1+t^2)}={\int }_{\alpha }^{+\propto }\frac{dt}{t(1+t^2)}$$$$but\frac{1}{t(1+t^2)}=\frac{1}{t}-\frac{t}{1+t^2}\Rightarrow \int \frac{dt}{t(1+t^2)}=ln/t/-\frac{1}{2}ln(1+t^2)+k$$$$=ln/\frac{t}{\sqrt{1+t^2}}/so{\int }_{\alpha }^{+\propto }\frac{dt}{t(1+t^2)}={\left[ln/\frac{t}{\sqrt{1+t^2}}/\right]}_{\alpha }^{+\propto }$$$$=-ln/\frac{\alpha }{\sqrt{1+{\alpha }^2}}/$$$$I=artan\left(\alpha \right)-\alpha ln/\frac{\alpha }{\sqrt{1+{\alpha }^2}}/.$$$$$$

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