# Math Help - Difficult Integral

1. ## Difficult Integral

Evaluate this integral:

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

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

I need a substitution $u=24x - 9 - 4x^2$ where $\frac{du}{dx} = 24 - 8x$.
Hence $6x+ 4 \equiv 14x - 20 + (24 - 8x)$.

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

So the integral can be written as:

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

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.

2. Try something like

$\frac{-6}{8}(-8x+24)+22=6x+4$ and then you can break it in two integrals and you will find a $arcsin$ function