# Thread: (Integral) making the numerator = 1

1. ## (Integral) making the numerator = 1

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

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

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

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

2. Originally Posted by Archduke01
$\int \frac {5x-6}{x^2-4x+29} dx$

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

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

But I'm confused as to how to do that.
First write the numerator, 5x- 6 as 5(x- 2)+ 4 so the integral is
$\int \frac{5(x-2)+ 4}{(x-2)^2+ 25} dx$.

Now separate it into two integrals:
$5\int \frac{x-2}{(x-2)^2+ 25}dx+ 4 \int\frac{1}{(x-2)^2+ 25} dx$.

3. Originally Posted by HallsofIvy
First write the numerator, 5x- 6 as 5(x- 2)+ 4
Is there some process you used to come up with that rewrite or is it something I'll simply have to have a feel for? I understand your following steps though, thanks a lot.