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