Integration by parts will help here. Letting \(u=\arctan x\) and \(\mathrm dv=\mathrm dx\), you end up with \(\mathrm du=\dfrac{\mathrm dx}{1+x^2}\) and \(v=x\). Now
\(\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du\)
\(\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx\)
For the remaining integral, setting \(y=1+x^2\) gives \(\dfrac{\mathrm dy}2=x\,\mathrm dx\), so
\(\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C\)
Putting everything together, you end up with
\(\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C\)
