integral

• Sep 12th 2009, 06:33 AM
alex23
integral
Hi, I have a problem with that:
$
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} dx
$
• Sep 12th 2009, 06:50 AM
CaptainBlack
Quote:

Originally Posted by alex23
Hi, I have a problem with that:
$
\int\limits_{0}^{1}\frac{3x - 4}{x^2 - 3x + 2}
$

Slight problem, the integrand has a nonintegrable singularity at one end of the range of integration!

CB
• Sep 12th 2009, 07:35 AM
alex23
oh - sry in previous integral was a mistake.

$
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} = (3*\frac{1}{2}*x^2 - 4x) |_{-1}^{0} - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)
$

$
*(-(x^2-3x+2)^{-2}) =
$

good for now ?
• Sep 12th 2009, 09:02 AM
songoku
Hi alex23
Quote:

Originally Posted by alex23
oh - sry in previous integral was a mistake.

$
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} = (3*\frac{1}{2}*x^2 - 4x) |_{-1}^{0} - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)
$

$
*(-(x^2-3x+2)^{-2}) =
$

good for now ?

I don't think your work is right. Do you know partial fraction ?
And you forgot "dx" in your question :)
• Sep 12th 2009, 10:50 AM
alex23
$
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} dx = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} dx =
$

$
(-5\frac{1}{2}) - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)*(-(x^2-3x+2)^{-2}) dx =
$

what is wrong ?
• Sep 12th 2009, 10:58 AM
skeeter
$\frac{3x-4}{x^2-3x+2}
$

$\frac{3x-4}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}
$

$3x-4 = A(x-2) + B(x-1)$

let $x = 2$ ... $B = 2$

let $x = 1$ ... $A = 1$

$\int \frac{1}{x-1} + \frac{2}{x-2} \, dx = \ln|x-1| + 2\ln|x-2| + C
$
• Sep 12th 2009, 11:42 AM
alex23
I remember that - this is definite integral.

But I think I have not definite integral. I have tried to solve this problem using this pattern:

$
\int\limits_{a}^{b} f'(x)g(x) dx = f(x)g(x) |_{a}^{b} - \int\limits_{a}^{b} f(x)g'(x) dx
$

But You said mine solutions from previous posts based on this pattern were wrong.

So now I have solution by skeeter - what next ? As I understand this is some other method based on definite integral and now I can't use above pattern ? So what should I do now ?
• Sep 12th 2009, 11:59 AM
skeeter
$\int_{-1}^0 \frac{1}{x-1} + \frac{2}{x-2} \, dx =$

$\left[\ln|x-1| + 2\ln|x-2|\right]_{-1}^0 =
$

$[0 + 2\ln{2}] - [\ln{2} + 2\ln{3}] = \ln{2} - \ln{9} = \ln\left(\frac{2}{9}\right)
$
• Sep 12th 2009, 02:25 PM
alex23
thx skeeter - Your solution is different than solution which I know and Your solution is easier :)

As I understand in Your solution we must count definite integral and then $|_{a}^{b}$

? Only that ?

So maybe mine pattern is for count area under curve ? I had integrals 3. years ago and I remember nothing.... :/ :(
• Sep 12th 2009, 08:16 PM
songoku
Hi alex23
Quote:

Originally Posted by alex23
$
\int\limits_{a}^{b} f'(x)g(x) dx = f(x)g(x) |_{a}^{b} - \int\limits_{a}^{b} f(x)g'(x) dx
$

But You said mine solutions from previous posts based on this pattern were wrong.

Your formula is right. It' integration by part. But your work is not right.
What are your f(x), f('x), g(x), and g'(x) ? :)