# Integrate 1/x^2+80x+1600

1. Mar 21, 2015

### Daniel Tyler

Hi,

I'm v rusty on my integration so was wondering could anyone give me some guidance on how to find the solution to the following integral

integral (1/x^2+80x+1600)dx between the limits x=80 and x=0

Any help is much appreciated!

Regards

Daniel

2. Mar 21, 2015

### Daniel Tyler

to be clearer its 1/(x^2+80x+1600)dx!

3. Mar 21, 2015

### Staff: Mentor

I's start by factoring the denominator. What is the context of the question? Where did this integral come up?

4. Mar 22, 2015

### Svein

To me, it looks like $\frac{1}{(x+40)^{2}}$. So, if you set $u=(x+40)$, you get $du=dx$ and $x=u-40$. This leaves you with an eminently solvable integral.