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What is the partial fraction decomposition of 6x+17/(x+1)^2?

Answer: A

What is the partial fraction decomposition of 6x+17/(x+1)^2? Answer: A-example-1
User Estebro
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2 Answers

5 votes

Explanation:


(6x+17)/((x+1)^2)

Since we have a repeated root (squared), we need two fractions:


(6x+17)/((x+1)^2)=(A)/(x+1)+(B)/((x+1)^2)

To find constants A and B, combine the fractions using common denominator.


(6x+17)/((x+1)^2)=(A(x+1))/((x+1)^2)+(B)/((x+1)^2)\\(6x+17)/((x+1)^2)=(Ax+A+B)/((x+1)^2)

Match the coefficients of the numerators:


A=6, A+B=17

Solve:


A=6, B=11

Therefore:


(6x+17)/((x+1)^2)=(6)/(x+1)+(11)/((x+1)^2)

User Alina Danila
by
8.0k points
1 vote

Answer:

Hi,

Answer 1

Explanation:


(6x+17)/((x+1)^2) =(A)/(x+1) +(B)/((x+1)^2) \\\\A(x+1)+B=6x+17\\\\Ax+A+B=6x+17\\\\A=6\\B=17-A\\B=11\\\\\boxed{(6x+17)/((x+1)^2) =(6)/(x+1) +(11)/((x+1)^2) }

Answer 1

User Mike Feng
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8.2k points

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