# Thread: Evaluate the Integral (Partial Fractions)

1. ## Evaluate the Integral (Partial Fractions)

For the first problem, this is how far I get:
http://www.rustybubble.net/mathwork.jpg
(big image)

But for the second, I don't know how to set it up.

Thank you for your help!

2. Originally Posted by birdz

For the first problem, this is how far I get:
http://www.rustybubble.net/mathwork.jpg
(big image)

But for the second, I don't know how to set it up.

Thank you for your help!
For the second problem you need a fraction for each degree of $(x-2)^2$.

So you need:

$\frac{A}{2x+1} + \frac{C}{x-2} + \frac{Dx+F}{(x-2)^2}
$

Good luck!

3. in the first integral you have two mistake the first one to decide A ,B ,C values cuz

x^2(x-2)=a/x+b/x^2+c/(x-2)

ax(x-2)+b(x-2)+cx^2=1
Ax^2-2Ax+Bx-2B+Cx^2=1
A+c=0
B-2A=0
-2B=1
so
B=-1/2 , A=-1/4 , C=1/4

this is the first one the second
the integrate of 1/x^2=-x^-1 not ln(x^2)

thats all

4. for repeated linear factors ...

$\frac{3x^2-9x+26}{(2x+1)(x-2)^2} = \frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}
$

$3x^2-9x+26 = A(x-2)^2 + B(2x+1)(x-2) + C(2x+1)$

let $x = 2$ ...

$20 = 5C$

$C = 4$

let $x = -\frac{1}{2}$ ...

$\frac{125}{4} = A \cdot \frac{25}{4}$

$A = 5$

$3x^2-9x+26 = 5(x-2)^2 + B(2x+1)(x-2) + 4(2x+1)$

let $x = 0$ ...

$26 = 20 - 2B + 4$

$B = -1$

$\frac{3x^2-9x+26}{(2x+1)(x-2)^2} = \frac{5}{2x+1} - \frac{1}{x-2} + \frac{4}{(x-2)^2}
$