# What is wrong with this method? (partial fractions)

1. Jan 31, 2012

### skyturnred

1. The problem statement, all variables and given/known data

So I am trying to integrate the following. I believe I have to use partial fractions.

$\frac{1}{(x+4)^{2}(x-3)}$

2. Relevant equations

3. The attempt at a solution

So I am trying to expand the above into something that I can integrate. So I equated the integrand with the following:

integrand = $\frac{A}{(x-3)}$+$\frac{Bx+C}{(x^{2}+16+8x) <--just the expanded form of (x+4)^{2}}$

So by creating a common denominator you get

1= A(x$^{2}$+16+8x)+(Bx+C)(x-3)

By inputing x=3 first, I solve for A=1/49
Then by imputing x=0 I get C=-33/49
And then by imputing x=1 I get B=-45/49

So plugging that into the original, I get:

integrand = $\frac{1}{49(x-3)}$-$\frac{(45/49)x-(33/49)}{(x+4)(x+4)}$

When I check my answer by plugging a random value of x into the integrand, and then into the RHS, I find that they are not equal. Where did I mess up?

2. Jan 31, 2012

### HallsofIvy

I don't know why you multiplied out $(x+ 4)^2$. If you leave it as is, you have
$$\frac{1}{(x+4)^2(x-3)}= \frac{A}{x-3}+ \frac{Bx+ C}{(x+4)^2}$$
Multiplying on both sides by $(x-3)(x+4)^2$ we get
$$1= A(x+4)^2+ (Bx+ C)(x- 3)$$

Now, you can take x= 3 and get $1= A(7)^2$ so A= 1/49 as you say.

If you let x= 0 you get $1= 16A+ C(-7)$ and knowing that A= 1/49, you get 7C= 16/49- 1= -33/49 so C= -33/343, not -33/49.

Finally, if you let x=-4, you get $1= (-4B+ C)(-7)$ and can solve for B.

3. Jan 31, 2012

### skyturnred

OK Thanks for the help!