# Integral help needed

## Homework Statement

$$\int \frac{26 dx}{{(169x^2+1)}^2}$$ <= the whole denominator is supposed to be squared...

## The Attempt at a Solution

So I converted the thing in the denominator so that it has a square root:
$$\int \frac{26 dx}{{\sqrt{169x^2+1}}^4}$$

Looking at the denominator, I realized I should do an inverse substitution:
13x=tan(t)
dx=sec2(t)*dt/13

I subbed that in to the equation before and got:
$$\int \frac{26 * sec^2(t) * dt}{13*sec^4(t)}$$

Simplifying which, I get:
2$$\int cos^2(t) dt$$

Then I tried doing integration by parts, but I got nowhere - I kept getting cos^2 again... Please help me, this question frustrates me. Thanks in advance!

(Sidenote: I finally got my formulas all pretty, yay!)