1. ## Integration problem

I am having a problem with this calculus problem:

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

3. Two more methods:

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

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