What is wrong with this method? (partial fractions)

[skyturnred]

Homework Statement

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

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

Homework Equations

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?

 

[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.

 

[skyturnred]

OK Thanks for the help!