Trigonometric integral (arctg)

Question

I have a problem with this integral. $$\int \text{arctan}(x-2)dx =\text{ }?$$ I tried integration by parts but it doesn't lead to right result.

Answer 1

To make the manipulations simpler we first substitute $x-2=u$ and this means $du=dv$ and we are left with $$\int\arctan{(x-2)}dx=\int\arctan{u}du$$ Now we integrate by parts $v=\arctan{u}$ and $du$. This yields $dv=\frac{1}{u^2+1}du$ and $u$. We have therefore $$\int\arctan{u}du=u\arctan{u}-\int\frac{u}{u^2+1}du$$ and with a substitution $w=u^2+1$ we are done