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

When I expand this, I can't eliminate to get one variable to get the variable.

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- Jun 9th 2010, 12:29 AMCthulHelp Partial Fractions
$\displaystyle \frac{1+x}{x(1+x^{2})}$

When I expand this, I can't eliminate to get one variable to get the variable.

- Jun 9th 2010, 12:37 AMProve It
Write it as

$\displaystyle \frac{A}{x} + \frac{Bx + C}{1 + x^2} = \frac{1 + x}{x(1 + x^2)}$

$\displaystyle \frac{A(1 + x^2) + (Bx + C)x}{x(1 + x^2)} = \frac{1 + x}{x(1 + x^2)}$

$\displaystyle A(1 + x^2) + (Bx + C)x = 1 + x$

$\displaystyle A + Ax^2 + Bx^2 + Cx = 1 + x$

$\displaystyle (A + B)x^2 + Cx + A = 0x^2 + x + 1$.

Equating coefficients of like powers of $\displaystyle x$ gives

$\displaystyle A + B = 0, C = 1, A = 1$.

So $\displaystyle B = -1$.

Therefore:

$\displaystyle \frac{1}{x} + \frac{1 - x}{1 + x^2} = \frac{1 + x}{x(1 + x^2)}$. - Jun 9th 2010, 01:47 AMCthul
Thank you, problem solved. I understand how now.