1. ## Integration problem

I am having a problem with this calculus problem:

$\displaystyle \int \frac{1}{y^2-16} dy$

I factored it to: $\displaystyle \int \frac{1}{(y-4)(y+4)} dy$

Then I tried integration by parts: $\displaystyle u=(y-4)^{-1} dv=(y+4)^{-1}$

$\displaystyle du=-(y-4)^{-2} v=ln(y+4)$

$\displaystyle \frac{ln(y+4)}{(y-4)}-\int \frac{ln(y+4)}{(y-4)^2} dy$

So I'm stuck, and I'm just wondering what is the best way to do this?
Peter

2. Originally Posted by flybynight
I am having a problem with this calculus problem:

$\displaystyle \int \frac{1}{y^2-16} dy$

I factored it to: $\displaystyle \int \frac{1}{(y-4)(y+4)} dy$

Then I tried integration by parts: $\displaystyle u=(y-4)^{-1} dv=(y+4)^{-1}$

$\displaystyle du=-(y-4)^{-2} v=ln(y+4)$

$\displaystyle \frac{ln(y+4)}{(y-4)}-\int \frac{ln(y+4)}{(y-4)^2} dy$

So I'm stuck, and I'm just wondering what is the best way to do this?
Peter
use the method of partial fractions ...

$\displaystyle \displaystyle \frac{1}{y^2-16} = \frac{A}{y+4} + \frac{B}{y-4}$

solve for $\displaystyle A$ and $\displaystyle B$ , then integrate each term individually

3. Two more methods:

2. Put $\displaystyle y = \tanh{t}$

3. Let $\displaystyle u = \frac{y-4}{y+4}$.