How do you express $1 / ((x+5)^2 (x-1))$ in partial fractions?

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Answer 1 · 2016-09-30T23:08:15

$1/((x+5)^2(x-1)) = -1/(6(x+5)^2) - 1/(36(x+5)) + 1/(36(x-1))$

$1/((x+5)^2(x-1)) = A/(x+5)^2 + B/(x+5) + C/(x-1)$

Multiplying both sides of this equation by $(x+5)^2$ we get:

$1/(x-1) = A+B(x+5)+(C(x+5)^2)/(x-1)$

Now let $x = -5$ to find:

$A = 1/((-5)-1) = -1/6$

Multiplying both sides of the first equation by $(x-1)$ we get:

$1/(x+5)^2 = (A(x-1))/(x+5)^2 + (B(x-1))/(x+5) + C$

Now let $x=1$ to find:

$C = 1/((1)+5)^2 = 1/36$

Going back to the first equation and making a common denominator we find:

$1/((x+5)^2(x-1)) = A/(x+5)^2 + B/(x+5) + C/(x-1)$

$color(white)(1/((x+5)^2(x-1))) = (A(x-1)+B(x-1)(x+5)+C(x+5)^2)/((x+5)^2(x-1))$

So equating the coefficients of $x^2$ in the numerator, we find:

$B = -C = -1/36$

How do you express 1 / ((x+5)^2 (x-1)) 1 / ((x+5)^2 (x-1)) in partial fractions?