Question

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

Answer: A

Answer 1

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)`

Answer 2

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