$\displaystyle \int \frac {5x-6}{x^2-4x+29} dx$

I complete the square and get; $\displaystyle \int \frac {5x-6}{(x-2)^2 + 25} dx$

If I can make the numerator equal to 1, I can use the following rule; $\displaystyle \int \frac {1}{x^2+a^2} dx = \frac {1}{a} arctan \frac {x}{a} + C$

But I'm confused as to how to do that.