Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 145358 by qaz last updated on 04/Jul/21

$$\mathrm{Evaluate}::{\int }_0^1\ln (1+{\mathrm{x}}^2)\cdot \arctan \left(\mathrm{x}\right)\mathrm{dx}=?$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$\mathrm{I}={\int }_0^1\ln (1+x^2)\arctan \left(x\right)dx$$$$\mathrm{I}={\left[x\ln (1+x^2)\arctan \left(x\right)\right]}_0^1$$$$-{\int }_0^{1}x(\frac{2x}{1+x^2}\arctan \left(x\right)+\frac{\ln (1+x^2)}{1+x^2})dx$$$$\mathrm{I}=\frac{\pi }{4}\ln \left(2\right)-2{\int }_0^1\arctan xdx$$$$+2{\int }_0^1\frac{\arctan x}{1+x^2}dx+{\int }_0^1\frac{1}{2}.\frac{2x}{1+x^2}.\ln (1+x^2)dx$$$$\mathrm{I}=\frac{\pi }{4}\ln \left(2\right)-2{[x\arctan x-\frac{1}{2}\ln (1+x^2)]}_0^1$$$$+2{\left[\frac{1}{2}{\arctan }^{2}x\right]}_0^1+\frac{1}{4}{\left[{\ln }^2(1+x^2)\right]}_0^1$$$$\mathrm{I}=\frac{\pi }{4}\ln \left(2\right)-2(\frac{\pi }{4}-\frac{1}{2}\ln \left(2\right))$$$$+\frac{{\pi }^2}{16}+\frac{1}{4}{\ln }^2\left(2\right)$$$$\mathrm{I}=\frac{{\pi }^2}{16}+\frac{\pi }{4}(\ln \left(2\right)-2)+\frac{1}{4}{\ln }^2\left(2\right)+\ln \left(2\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com