Partial Fractions

• Jul 9th 2008, 10:26 AM
JonathanEyoon
Partial Fractions
I'm not certain why but I can't seem to get the right answer no matter how many times I run through this

Integrate (x^2) / ((x - 3)(x + 2)^2)

(A / x - 3) + (B / x + 2) + (C / (x + 2)^2)

My values for A and C seem correct from the method I was taught by the professor

A = 9 / 25 C = - 4 / 5

Now to get B, I equated for cubics.

x^2 = ((9/25)(x + 2)^3) + B(x - 3)(x + 2)^2 - (4/5)(x - 3)(x + 2)

Now the B I got was - 9 / 25.

Can someone tell me if I've done this right so far? Thanks
• Jul 9th 2008, 10:45 AM
galactus
You didn't multiply by $(x-3)(x+2)(x+2)^{2}$ did you?.

Should use $(x-3)(x+2)^{2}$

$(x+2)^{2}(x-3)\cdot\frac{A}{x-3}+(x+2)^{2}(x-3)\cdot\frac{B}{x+2}+(x+2)^{2}(x-3)\cdot\frac{C}{(x+2)^{2}}$

Cancel appropriately, expand out, and we get:

$Ax^{2}+4Ax+4A+Bx^{2}-Bx-6B+Cx-3C=x^{2}$

Equate coefficients:

$A+B=1$

$4A-B+C=0$

$4A-6B-3C=0$

Solve the system and we get:

$A=\frac{9}{25}, \;\ B=\frac{16}{25}, \;\ C=\frac{-4}{5}$

So, we get:

$\frac{9}{25(x-3)}+\frac{16}{25(x+2)}-\frac{4}{5(x+2)^{2}}$
• Jul 9th 2008, 10:49 AM
JonathanEyoon
Yea I did do that (Headbang). Thanks for the help I appreciate it.
• Jul 9th 2008, 10:55 AM
Jhevon
Quote:

Originally Posted by JonathanEyoon
I'm not certain why but I can't seem to get the right answer no matter how many times I run through this

Integrate (x^2) / ((x - 3)(x + 2)^2)

(A / x - 3) + (B / x + 2) + (C / (x + 2)^2)

My values for A and C seem correct from the method I was taught by the professor

A = 9 / 25 C = - 4 / 5

Now to get B, I equated for cubics.

x^2 = ((9/25)(x + 2)^3) + B(x - 3)(x + 2)^2 - (4/5)(x - 3)(x + 2)

Now the B I got was - 9 / 25.

Can someone tell me if I've done this right so far? Thanks

Alternatively,

$\frac {x^2}{(x - 3)(x + 2)^2} = \frac A{x - 3} + \frac B{x + 2} + \frac C{(x + 2)^2}$

multiply through by $(x - 3)(x + 2)^2$

$\Rightarrow x^2 = A(x + 2)^2 + B(x - 3)(x + 2) + C(x - 3)$

Let $x = -2$
$\Rightarrow 4 = -5C \implies \boxed{C = - \frac 45}$

Let $x = 3$
$\Rightarrow 9 = 25A \implies \boxed{A = \frac 9{25}}$

Let $x = 0$ ....just some random value, 0 is easy to work with:
$\Rightarrow 0 = 4A - 6B - 3C$

plug in the values for $A$ and $C$ and solve for $B$, we get: $\boxed{B = \frac {16}{25}}$. as the great galactus said
• Jul 9th 2008, 11:23 AM
JonathanEyoon
Thanks Jhevon for the alternative method!! I'll remember that