Author Topic: Partial fracs (Read 651 times) Tweet Share

fredrick · « **on:** November 04, 2008, 01:29:19 pm »

Given that $\frac{10x}{(x^2+1)(3x+1)}$ can be written in the form $\frac{x+A}{x^2+1} + \frac{B}{3x+1}$ find A and B, show working plz

ell · « **Reply #1 on:** November 04, 2008, 01:47:22 pm »

$10x \equiv (x+A)(3x+1) + B(x^{2}+1)$

Let $x=-\frac{1}{3}$

$\implies -\frac{10}{3}=\frac{10B}{9}$

$\implies B=-3$

So $10x \equiv (x+A)(3x+1) -3(x^{2}+1)$

Now expand:

$10x \equiv 3x^2 + x + 3Ax + A -3x^{2} -3$

$10x \equiv (3A +1)x + A -3$

Equate coefficients:

$3A+1=10$ (1)

$A-3=0$ (2)

So $A=3$

$\therefore \frac{10x}{(x^2+1)(3x+1)} \equiv \frac{x+3}{x^2+1} - \frac{3}{3x+1}$

fredrick · « **Reply #2 on:** November 04, 2008, 02:18:32 pm »

thanks

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Author Topic: Partial fracs (Read 651 times) Tweet Share

fredrick

Partial fracs

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