# Help with Solid of Revolution/Partial Fraction Decompisition Question

1. Feb 24, 2014

### student93

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

Question is attached in this post.

2. Relevant equations

Question is attached in this post.

3. The attempt at a solution

Disk Method with the radius equal to 1/(x^2+5x+6)

∫A/(x+2) + B/(x+2)^2 + C/(x+3) + D/(x+3)^2 from 0 to 3

I've used algebra to solve for both B and D, both of which are equal to 1, but can't seem to be able to solve for the values of A and C.

The answer to the question is 2π(5ln2 - 2ln5)

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2. Feb 24, 2014

### LCKurtz

3. Feb 24, 2014

### student93

1=A(x+2)(x+3)^2 + B(x+3)^2 + C(x+2)^2(x+3) +D(x+2)^2

I first let x=-3, and was able to get D=1

Next I let x=-2 and was able to get B=1

However, I can't seem use this method to get the values of either A or C.

4. Feb 24, 2014

### pasmith

That's because $(x + 2)(x + 3)$ vanishes when $x = -3$ or $x = -2$, so to make the coefficients of $A$ and $C$ not vanish you need to choose some other values for $x$.

That will leave you with two equations in two unknowns once you substitute the known values of $B$ and $D$.

5. Feb 24, 2014

### LCKurtz

Try plugging in some other values for $x$ to get more equations. Alternatively you can compare coefficients of various powers of $x$ on both sides. For example, maybe try $x=0$ or looking at the coefficient of $x^3$ on both sides.

6. Feb 24, 2014

### student93

Could you clarify exactly what you mean by comparing coefficients of various powers of x on both sides?

Also, alternatively, can I use Trig Substitution instead to solve this problem?

7. Feb 24, 2014

### LCKurtz

You have $1=A(x+2)(x+3)^2 + B(x+3)^2 + C(x+2)^2(x+3) +D(x+2)^2$. It's pretty easy to see the coefficient of $x^3$ on the right side is $A+C$ and you have $0x^3$ on the left so $A+C = 0$.

I don't see how trig substitution would be useful nor why you would want to try it.