Partial Fractions

1. Mar 26, 2008

[SOLVED] Partial Fractions

1. Evaluate:

$$\int \frac{dx}{x^{2} -1}$$

Attempt:

$$\int \frac{dx}{x^{2} -1}$$

$$= \int \frac{dx}{(x+1)(x-1)}$$

$$= \frac{A}{(x+1)} + \frac{B}{(x-1)}$$

$$= \frac{Ax - A + Bx + B}{(x+1)(x-1)}$$

Where do I got from here? Thanks

2. Mar 26, 2008

tiny-tim

(erm … what happened to the integral sign, and dx … ?)

Easy! You just solve Ax - A + Bx + B = 1.

So A = … , B = … ?

3. Mar 26, 2008

so A = Ax + Bx + B -1
and B = -Ax + A - Bx -1??

4. Mar 26, 2008

rock.freak667

Ax-A+Bx+B=1 for all values of x

meaning for any value of x you put in there, it will always be equal to 1
Choose some suitable values for x to get A and B

5. Mar 26, 2008

Snazzy

You could do A(x-1) + B(x+1) = 1

Then to get rid of the A, substitute x = 1 and find B.

To get rid of the B, substitute x = -1 and find A.

6. Mar 26, 2008

That means:

(1/2)ln(x-1) - (1/2)ln(x+1) + C

Therefore Since: (1/2)ln(x-1) = ln(x-1)^1/2 and (1/2)ln(x +1)= (1/2)ln(x +1)^1/2

ln(x-1)^1/2 - (1/2)ln(x +1)^1/2 + C

Natural log rule ln a - ln b = ln (a/b)

Thank you =)

7. Mar 27, 2008

tiny-tim

… never substitute …

No!

It's a polynomial, and you just put all the coefficients equal to 0.

It's (A+B)x + (B-A-1) = 0,

so both the brackets must be 0.

Never substitute - it makes it look as if you don't understand what a polynomial is!

(I agree that, technically, that's the same as substituting x = 0 and ∞; but that requires an extra line, and it doesn't work if you have x^2 or higher.)

8. Mar 27, 2008

HallsofIvy

Staff Emeritus
Tiny Tim, since that equation is true for all x, you certainly could get substitute any two values of x to get 2 equations to solve for a and b. Of course, it is true that if a polynomial is 0 for all x, then its coefficients must all be 0- that's typically simpler than substituting specific values for x.

Ladies Man, a somewhat simpler method for partial fraction is this:
You want to find A, B such that
$$\frac{A}{x+1}+ \frac{B}{x-1}= \frac{1}{x^2- 1}$$
Multiply both sides by x2-1 to eliminate the denominators:
[tex]A(x-1)+ B(x+ 1)= 1[/itex]
Now substitute x= 1 and x= -1 to find A and B.

9. Mar 27, 2008