# Math Help - integration

1. ## integration

how do u do this integration? ive tried so many times and it just doesnt work!! i tried the completing the square way but i didnt get the right answer.

integral xdx/(x^2-2x+3)

2. Notice that $\int \frac{x}{x^2-2x+3} = \int \frac{x}{(x-1)^2+2}$

Sub u = x-1 and you'll get

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

Splitting it will get you two simpler integrals that I think you should see right away.

3. Originally Posted by chukie
how do u do this integration? ive tried so many times and it just doesnt work!! i tried the completing the square way but i didnt get the right answer.

integral xdx/(x^2-2x+3)
$\frac{x}{x^2-2x+3} = \frac{x-1}{x^2-2x+3} + \frac{1}{x^2-2x+3}$

Now, $\int \frac{x-1}{x^2-2x+3} dx = \frac{1}{2}\int \frac{(x^2-2x+3)'}{x^2-2x+3}dx$ (Think logarithm here).

And, $\int \frac{dx}{x^2 - 2x+3} = \int \frac{dx}{(x-1)^2 + (\sqrt{2})^2} dx$ (Think arctangent).