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