# Thread: [SOLVED] Partial Fraction Integration

1. ## [SOLVED] Partial Fraction Integration

Evaluate the integral of 4/(1-x)(1+x)^2. the answer is supposed to be -2/x+1 + ln |x+1/x-1|

I did A/1-x + B/1+x + C/1+x)^2 and eventually got this after multiplying by common denominators: -Bx^3+x^2(A-B)+x(2A+B-C)+A+B+C=1 and do not know how to solve for A, B, or C. Am I doing this right?

2. This is too difficult to follow. Something like this would ideally be in Latex, but if you can't code it you have got to be clear with parentheses. It's a headache to look at, and that will discourage people from helping you. So if you clean it up I'll take another look. Maybe also include an intermediate steps. Again this makes things easier to follow and easier to spot errors.

3. sorry it's a mess...How do you do math code?

4. ## Partial Fractions

Hello juicysharpie
Originally Posted by juicysharpie
Evaluate the integral of 4/(1-x)(1+x)^2. the answer is supposed to be -2/x+1 + ln |x+1/x-1|

I did A/1-x + B/1+x + C/1+x)^2 and eventually got this after multiplying by common denominators: -Bx^3+x^2(A-B)+x(2A+B-C)+A+B+C=1 and do not know how to solve for A, B, or C. Am I doing this right?
$\frac{4}{(1-x)(1+x)^2} = \frac{A}{1-x}+\frac{B}{1+x}+\frac{C}{(1+x)^2}$

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

$\Rightarrow A(1+x)^2+B(1-x)(1+x)+C(1-x) \equiv 4$

$\Rightarrow A(1+2x+x^2)+B(1-x^2)+C(1-x) \equiv 4$

$\Rightarrow x^2(A-B) + x(2A-C)+A+B+C \equiv 4$

Compare coefficients:

$A-B=0$

$2A-C=0$

$A+B+C=4$

$\Rightarrow A=B=1, C=2$

Can you take it from here?

PS If you want to see the Latex code behind any of these expressions, just click on them, and the code will open up in a separate window.

5. Thanks for the help Grandad!! It was very nice of you to answer both of my questions.