# Integral x/x^2-2x+3

• April 18th 2009, 08:34 PM
cammywhite
Integral x/x^2-2x+3
$\int \frac {x}{x^2-2x+3}dx$

The model answer is $\frac {ln(x^2-2x+3)}{2}+\frac{artan\frac{x+1}{\sqrt{2}}}{\sqrt{2 }}+C$

First step, do I make $u=x^2-2x+3$?
• April 18th 2009, 08:44 PM
TheEmptySet
Quote:

Originally Posted by cammywhite
$\int \frac {x}{x^2-2x+3}dx$

The model answer is $\frac {ln(x^2-2x+3)}{2}+\frac{artan\frac{x+1}{\sqrt{2}}}{\sqrt{2 }}+C$

First step, do I make $u=x^2-2x+3$?

First start by completing the square in the denominator

$\int \frac {x}{(x-1)^2+2}dx$ Now let $u=x-1 \implies du=dx$

$\int \frac{u+1}{u^2+2}du=\int \frac{u}{u^2+2}du+\int \frac{1}{u^2+2}du$

You should be able to finish from here