1. ## Partial Fraction

$integrate: \frac{x^3 + 3x^2 + 5x -9}{(x-5)(x+1)^3} dx$

I worked it down to $x^3 + 3x^2 + 5x -9 = A(x+1)^3 + B(x-5)(x+1)^2 + c(x-5)(x+1) + D(x-5)$

and then

x = 5
a= 1

x = -1
-12 = b

whats confusing me is that if i let x = -1 for c i would get -12 = C0 which is -12 = 0
is my equation wrong?

2. Originally Posted by x5pyd3rx
$integrate: \frac{x^3 + 3x^2 + 5x -9}{(x-5)(x+1)^3} dx$

I worked it down to $x^3 + 3x^2 + 5x -9 = A(x+1)^3 + B(x-5)(x+1)^2 + c(x-5)(x+1) + D(x-5)$

and then

x = 5
a= 1

x = -1
-12 = b

whats confusing me is that if i let x = -1 for c i would get -12 = C0 which is -12 = 0
is my equation wrong?
Never thought about doing the question like that but lets roll with it...

If x=5, everything bu the A term is 0 so we get...

216A = 216.

So A = 1.

If x=-1, everything but the D term cancels and you get...

-6D = -12, D = 2

Now set x to be something generic like 0.

Chances are you'll get something like ...B + ...C = -9

Then set x to be something else. Like 1 for example.

Once again you'll get an equation involving B and C.

Solve for B and C using simultaneous equations.

3. Alternatively you could see that you'll get Ax^3 + ... + Bx^3 + ...

So B must be equal to 0.

Something tells me that won't work though. You sure that equation is going to work..?

4. Actually yeah it will be. C=0 as well.