Evaluate this integral:

$\displaystyle \int \frac{6x+4}{\sqrt(24x - 9 - 4x^2)} dx$

Following an example in my texbook, I've split the top line:

I need a substitution $\displaystyle u=24x - 9 - 4x^2$ where $\displaystyle \frac{du}{dx} = 24 - 8x $.

Hence $\displaystyle 6x+ 4 \equiv 14x - 20 + (24 - 8x)$.

Also, completing the square $\displaystyle 24x - 9 - 4x^2 = -4(x-3)^2 + 27 $

So the integral can be written as:

$\displaystyle \int \frac{14x}{\sqrt(24x - 9 - 4x^2)} dx - 20\int \frac{dx}{\sqrt(24x - 9 - 4x^2)} + \int \frac{24-8x}{\sqrt(24x - 9 - 4x^2)} dx $

I know I'll end up with an inverse trig function somewhere but that's where I get stuck.

Could someone please explain what to do. Thanks a lot.