Integrate by partial fractions

**Vamz** · February 9th 2011, 04:33 PM

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

What am I doing wrong here?

First, I long divide & get:

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

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

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

I figure that A = 1, B = 2

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

This should integrate to...

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

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

Thank you!

**harish21** · February 9th 2011, 04:38 PM

the answer is right!

Check it here.

**mr fantastic** · February 9th 2011, 04:45 PM

Originally Posted by Vamz

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

What am I doing wrong here?

First, I long divide & get:

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

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

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

I figure that A = 1, B = 2

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

This should integrate to...

$ 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.

**topsquark** · February 9th 2011, 04:48 PM

Originally Posted by Vamz

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

I must be getting picky in my old age....

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

-Dan

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