Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 27

$xarctanx-\frac{1}{2}\ln ({x^{2}+1})+C$

Given $\int arctanxdx$ Use intergration by parts. $u=arctanx,dv=\int dx$ $u=\frac{1}{x^{2}+1}dx,v=x$ $=xarctanx-\int \frac{x}{x^{2}+1}dx$ Let $u=x^{2}+1$ $\frac{du}{dx}=2x$ $dx=\frac{1}{2x}du$ $=xarctanx-\frac{1}{2}\int \frac{du}{u}$ $=xarctanx-\frac{1}{2}\ln u+C$ $=xarctanx-\frac{1}{2}\ln ({x^{2}+1})+C$