Find the partial fraction decomposition:

**Selim** · Mar 21st 2010, 01:24 PM

Find the partial fraction decomposition:

$\frac{3x+3}{x^2+3x}$

**skeeter** · Mar 21st 2010, 01:31 PM

Originally Posted by Selim

Find the partial fraction decomposition:

$\frac{3x+3}{x^2+3x}$

$\frac{3x+3}{x(x+3)} = \frac{A}{x} + \frac{B}{x+3}$

$3x+3 = A(x+3) + Bx$

solve for A and B

**Selim** · Mar 21st 2010, 01:36 PM

Is the answer simply $\frac{A}{x} + \frac{B+C}{x+3}$

Aw, thanks, I would have been incorrect. Thank you again, Skeeter.

So then using x=0 and x=-3 I found that A=1 and B=2.

**skeeter** · Mar 21st 2010, 01:43 PM

Originally Posted by Selim

Is the answer simply $\frac{A}{x} + \frac{Bx+C}{x+3}$ ... no

$3x+3 = A(x+3) + Bx$

let $x = 0$ ... $A = 1$

let $x = -3$ ... $B = 2$

$\frac{3x+3}{x(x+3)} = \frac{1}{x} + \frac{2}{x+3}$

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