How do you decompose $(3p-1)/(p^2-1)$ into partial fractions?

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Answer 1 · 2016-12-24T14:52:44

$p^2 - 1$ can be factored as $(p + 1)(p - 1)$ .

$A/(p - 1) + B/(p + 1) = (3p - 1)/((p + 1)(p - 1)$

$A(p + 1) + B(p - 1) = 3p - 1$

$Ap + A + Bp - B = 3P - 1$

$(A + B)p + (A - B) = 3P - 1$

We now write a system of equations.

${(A + B = 3), (A - B = -1):}$

$B = 3 - A -> A - (3 - A) = -1$
$A + A - 3 = -1$
$2A = 2$
$A = 1$

$1 + B = 3$

$B = 2$

Thus, the partial fraction decomposition is $1/(p - 1) + 2/(p + 1)$ .

Hopefully this helps!

How do you decompose (3p-1)/(p^2-1)(3p-1)/(p^2-1) into partial fractions?