1. ## integral

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

2. Originally Posted by alex23
Hi, I have a problem with that:
$\displaystyle \int\limits_{0}^{1}\frac{3x - 4}{x^2 - 3x + 2}$
Slight problem, the integrand has a nonintegrable singularity at one end of the range of integration!

CB

3. oh - sry in previous integral was a mistake.

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

good for now ?

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

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

good for now ?
I don't think your work is right. Do you know partial fraction ?
And you forgot "dx" in your question

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

what is wrong ?

6. $\displaystyle \frac{3x-4}{x^2-3x+2}$

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

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

let $\displaystyle x = 2$ ... $\displaystyle B = 2$

let $\displaystyle x = 1$ ... $\displaystyle A = 1$

$\displaystyle \int \frac{1}{x-1} + \frac{2}{x-2} \, dx = \ln|x-1| + 2\ln|x-2| + C$

7. 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:

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

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 ?

8. $\displaystyle \int_{-1}^0 \frac{1}{x-1} + \frac{2}{x-2} \, dx =$

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

$\displaystyle [0 + 2\ln{2}] - [\ln{2} + 2\ln{3}] = \ln{2} - \ln{9} = \ln\left(\frac{2}{9}\right)$

9. 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 $\displaystyle |_{a}^{b}$

? Only that ?

So maybe mine pattern is for count area under curve ? I had integrals 3. years ago and I remember nothing.... :/

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

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) ?