# Evaluating this integral

• Feb 9th 2009, 06:02 PM
juicysharpie
Evaluating this integral
The integral of 1/(x-3)(x+2)
The answer is 1/5 ln |x-2/x| + C

I split it up to be A/(x-3) + B/(x+2) = (Ax+2A+Bx-3B)/(x-3)(x+2) = (x(A+B)+2A-3B)/(x-3)(x+2). Then I set A+B = 0 so 2A-3B must equal 1? Am I doing this right? I don't exactly know where to start by solving for A or B. (Worried)

Thank you!
• Feb 9th 2009, 07:01 PM
kbzoon
Note that: $
\frac{1}{\left( x-3 \right)\left( x-2 \right)}=\frac{x-2-\left( x-3 \right)}{\left( x-3 \right)\left( x-2 \right)}=\frac{1}{\left( x-3 \right)}-\frac{1}{\left( x-2 \right)}
$

$
\Rightarrow \int{\frac{dx}{\left( x-3 \right)\left( x-2 \right)}}=\int{\frac{dx}{\left( x-3 \right)}}-\int{\frac{dx}{\left( x-2 \right)}}=\ln \left| x-3 \right|-\ln \left| x-2 \right|+C
$

$
\int{\frac{dx}{\left( x-3 \right)\left( x-2 \right)}}=\ln \left| \frac{x-3}{x-2} \right|+C
$

Another way to get a result.
• Feb 16th 2009, 06:28 PM
Nacho
Quote:

Originally Posted by juicysharpie
I set A+B = 0 so 2A-3B must equal 1? Am I doing this right?

That´s right

And you can begin solving for A or B, is same