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March 19th, 2017, 10:18 AM   #1
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Improper integral

I attached the problem below. My question is where does the 1/4 come from? I see that 1/((x+3)(x-1)) becomes (1/(x-1))-(1/(x+3)) because of the 1/4 we took out, but I don't understand this step. Would someone please clarify? Thank you!
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 March 19th, 2017, 10:30 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400 $\dfrac{1}{(x+3)(x-1)} = \dfrac{A}{x+3}+\dfrac{B}{x-1}$ $1 = A(x-1) + B(x+3)$ $x=1 \implies 1 = 4B \implies B = \dfrac{1}{4}$ $x=-3 \implies 1 = -4A \implies A = -\dfrac{1}{4}$ $\dfrac{1}{(x+3)(x-1)} = \dfrac{1}{4}\bigg[\dfrac{1}{x-1}-\dfrac{1}{x+3}\bigg]$

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