Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 31099 by abdo imad last updated on 02/Mar/18

$$find{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2x\right)-arctanx}{x}dx.$$

Commented by abdo imad last updated on 04/Mar/18

$$I={lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)/I\left(\xi \right)={\int }_0^{\xi }\frac{artan\left(2x\right)-arctanx}{x}dx$$$$I\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(2x\right)}{x}dx-{\int }_0^{\xi }\frac{arctanx}{x}dx$$$$ch.2x=tgive{\int }_0^{\xi }\frac{arctan\left(2x\right)}{x}dx={\int }_0^{2\xi }\frac{arctan\left(t\right)}{\frac{t}{2}}\frac{dt}{2}$$$$={\int }_0^{2\xi }\frac{arctant}{t}dt\Rightarrow I\left(\xi \right)={\int }_0^{2\xi }\frac{arctanx}{x}dx-{\int }_0^{\xi }\frac{arctanx}{x}dx$$$$={\int }_{\xi }^{2\xi }\frac{arctanx}{x}dxbut\mathrm{\exists }c\in ]\xi ,2\xi [/I\left(\xi \right)=artan\xi {\int }_{\xi }^{2\xi }\frac{dt}{t}$$$$=ln\left(2\right)arctan\xi \Rightarrow {lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)=\frac{\pi }{2}ln2\Rightarrow I=\frac{\pi }{2}ln\left(2\right).$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com