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 170725 by Sotoberry last updated on 29/May/22

Answered by FelipeLz last updated on 30/May/22

$$u={\tan }^{-1}\left(x\right)\Rightarrow du=\frac{1}{1+x^2}dx$$$$dv=1dx\Rightarrow v=x$$$$\int {\tan }^{-1}\left(x\right)dx$$$$\int {\tan }^{-1}\left(x\right)1dx=x{\tan }^{-1}\left(x\right)-\int \frac{x}{1+x^2}dx=x{\tan }^{-1}\left(x\right)-\frac{1}{2}\ln \mid 1+x^2\mid$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com