# Thread: Integrate by partial fractions

1. ## Integrate by partial fractions

$\displaystyle \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx$

What am I doing wrong here?

First, I long divide & get:

$\displaystyle \int 6x+4+\frac{3x-1}{x^2-1}dx$

$\displaystyle \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx$

$\displaystyle A(x-1)+B(x+1)=3x-1$

I figure that A = 1, B = 2

$\displaystyle \int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx$

This should integrate to...

$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$

this answer appears to be wrong...
Where did I go wrong?

Thank you!

Check it here.

3. Originally Posted by Vamz
$\displaystyle \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx$

What am I doing wrong here?

First, I long divide & get:

$\displaystyle \int 6x+4+\frac{3x-1}{x^2-1}dx$

$\displaystyle \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx$

$\displaystyle A(x-1)+B(x+1)=3x-1$

I figure that A = 1, B = 2

$\displaystyle \int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx$

This should integrate to...

$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$

this answer appears to be wrong...
Where did I go wrong?

Thank you!
I think you'll find that A = 2 and B = 1.

4. Originally Posted by Vamz
$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$
I must be getting picky in my old age....

The solution is
$\displaystyle 3x^2+4x+2ln|x+1| + ln|x-1| + C$

-Dan