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