Partial Fractions: Solve Integral of 1/y^2-1 dx

[The_ArtofScience]

Hi!

There's this one problem that I'm having troubles with. I've tried using the decomposition method, but I've ended up getting a messy answer. If someone can give me tips or the solution to the problem, I'll appreciate it. Here's the problem: solve the integral of 1/ y^2-1 dx.

 

[nicksauce]

I assume you mean $$\int\frac{1}{x^2-1}dx$$.

So if you want to use partial fractions, you can decompose it as such:
$$\frac{1}{x^2-1}=\frac{A}{x+1}+\frac{B}{x-1}$$. How would you continue?

 

[The_ArtofScience]

I used that method however it came out pretty ugly.

(1) A/ y+1 + B/ y-1

(2) 1 = (y+1) (y-1) A/ y+1 + (y+1) (y-1) B/ y-1 -I multiplied each side by the LCD

(3) 1 = A(y-1) +B(y+1) -Cancelled algebraically

(4) 1 = y(A+B) + (-A+B)

Since there's only one constant, am I assume that (A+B) =1 and similarly that -A+B =1? In order to identify A and B, I set up the system of equations as:

A+B =1
-A+B= 1

I know there's something wrong here because for any subsitution I would get "0" Plus, I've never seen a partial decomp problem where A and B would equal to the same number.

 

[nicksauce]

A little mistake at the end. Since there is no y term on the left side, you require A+B = 0, while -A + B = 1.

 

[The_ArtofScience]

Thanks Nicksauce,

The only rational answer I can see why A+B can equal to 0 is because the costants on both sides of the equation are equal somehow

Here's what I got from using that helpful tip

(1) 1/ y^2-1 = A/ y-1 + B/ y+1

(2) -1/2/ y-1 + 1/2/ y+1

(3) integral 1/ y^2-1 dx = integral - dx/ 2(y-1) + dx/ 2(y+1)

(4) -1/2 integral dx/ y-1 + 1/2 integral dx/ y+1

= 1/2 ln (y+1) -1/2 ln (y-1) whew!

= 1/2 ln ((y-1)/(y+1))

Thanks for the help! :-)

 

[nicksauce]

No problemo.