Thread: Find the partial fraction decomposition:

Find the partial fraction decomposition:

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

Originally Posted by Selim

Find the partial fraction decomposition:

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

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

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

solve for A and B

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

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.

Originally Posted by Selim

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

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

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

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

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