$\displaystyle \int \frac {x}{x^2-2x+3}dx$
The model answer is $\displaystyle \frac {ln(x^2-2x+3)}{2}+\frac{artan\frac{x+1}{\sqrt{2}}}{\sqrt{2 }}+C$
First step, do I make $\displaystyle u=x^2-2x+3$?
First start by completing the square in the denominator
$\displaystyle \int \frac {x}{(x-1)^2+2}dx$ Now let $\displaystyle u=x-1 \implies du=dx$
$\displaystyle \int \frac{u+1}{u^2+2}du=\int \frac{u}{u^2+2}du+\int \frac{1}{u^2+2}du$
You should be able to finish from here