Find the partial fraction decomposition: $\displaystyle \frac{3x+3}{x^2+3x}$
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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.
Last edited by Selim; Mar 21st 2010 at 01:49 PM.
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 $\displaystyle x = -3$ ... $\displaystyle B = 2$ $\displaystyle \frac{3x+3}{x(x+3)} = \frac{1}{x} + \frac{2}{x+3}$
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