# Solving This Integral

• Dec 3rd 2008, 11:51 AM
trigoon
Solving This Integral
I'm really having trouble understanding how to solve the following integrals, hopefully someone can help me :) (Bow)

∫ (3/(x^2 - 7x +12))

I'm just not sure even how to start. I know that I can take out the constant and make it:

3 ∫ (1/(x^2 - 7x +12))

But what should I do after that, I know that my answer will involve utilizing ln
• Dec 3rd 2008, 12:12 PM
running-gag
Hi

You can start by rewriting
$\frac{1}{x^2-7x+12}$ in "simple elements"

using the fact that $x^2-7x+12 = (x-4)(x-3)$

$\frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}$

$\int \frac{3}{x^2-7x+12} \, dx = 3 \int (\frac{1}{x-4} - \frac{1}{x-3}) \, dx = 3\; ln\frac{|x-4|}{|x-3|}$
• Dec 3rd 2008, 12:21 PM
trigoon
Quote:

Originally Posted by running-gag
Hi

You can start by rewriting
$\frac{1}{x^2-7x+12}$ in "simple elements"

using the fact that $x^2-7x+12 = (x-4)(x-3)$

$\frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}$

$\int \frac{3}{x^2-7x+12} \, dx = 3 \int (\frac{1}{x-4} - \frac{1}{x-3}) \, dx = 3\; ln\frac{|x-4|}{|x-3|}$

Thanks! I had you until this part:
$\frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}$

How did you get that?
• Dec 3rd 2008, 12:36 PM
running-gag
Quote:

Originally Posted by trigoon
Thanks! I had you until this part:
$\frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}$

How did you get that?

Experience ... (Wink)

This is the nominal way to "decompose in simple elements" (translation from French, I do not know if it is called this way in English)

$\frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \,(\frac{1}{x-a} - \frac{1}{x-b})$
• Dec 3rd 2008, 01:34 PM
trigoon
Thanks a lot this really helps!