1. Partial fractions and integrals

Use the method of partial fractions to evaluate the following integral:
$\int (x^3 + 4x^2 - 6x)/(x^2 - 2x + 1)dx$

I have also solved for the denominator that $(x^2 - 2x + 1)$ gives $(x-1)^2$

Can anybody assist me to find A + B, for the numerators? Given the form:
$A/(x-1) + B/(x-1)$

Thank you and any help is much appreciated,

Dranalion

2. you can't apply it yet, you need to do a polynomial long division first, since the $\deg p(x)>\deg q(x)$ given $p(x)$ and $q(x)$ polynomials for numerator and denominator respectively.

3. Applying polynomial long division, I got the result:
$\int x + 6 + \frac{(5x - 6)}{(x-1)(x-1)} dx$

Given that earlier, I found that $x^2 - 2x + 1 = (x-1)(x-1)$

Now, how do I make the $\frac{(5x - 6)}{(x-1)(x-1)}$ part into the format:

$\frac{(A + B)}{(x-1)(x-1)}$, where I need to find A and B

4. no actually, the descomposition is $\frac{5x-6}{(x-1)(x-1)}=\frac{a}{x-1}+\frac{b}{(x-1)^{2}}.$

5. How do I find A + B, either way?

6. Originally Posted by Dranalion
How do I find A + B, either way?
The right hand side can be written $\frac{a(x-1) + b}{(x - 1)^2}$. Compare the numerator of the right hand side to the numerator of the left hand side.

7. I'm lost.

$5x - 6 = Ax - (A+B)$

=> $5 = A ...(1)$
=> $-6 = -A + B ...(2)$

Substituting A = 5 into (2) gives:

=> $-6 = -5 + B$

Which gives B = -1

Using my graphics calculator to check, I have somehow managed to get two different curves? As in $x + 6 + \frac{5}{x-1} - \frac{1}{x-1}$ $\frac{(x^3 + 4x^2 - 6x)}{(x^2 - 2x + 1)}$

8. Originally Posted by Dranalion
I'm lost.

$5x - 6 = Ax - (A+B)$

=> $5 = A ...(1)$
=> $-6 = -A + B ...(2)$

Substituting A = 5 into (2) gives:

=> $-6 = -5 + B$

Which gives B = -1

Using my graphics calculator to check, I have somehow managed to get two different curves? As in $x + 6 + \frac{5}{x-1} - \frac{1}{x-1}$ $\frac{(x^3 + 4x^2 - 6x)}{(x^2 - 2x + 1)}$

Of course they're not equal. Negligence begets error.

Substitute your values for A and B into the expression given to you in post #4.