Question

How do I find the partial-fraction decomposition of `(3x^2+2x-1)/((x+5)(x^2+1))`?

Answer 1

`(3x^2+2x-1)/((x+5)(x^2+1))` becomes ...

`(3x^2+2x-1)/((x+5)(x^2+1))=A/(x+5)+(Bx+C)/(x^2+1)`

Multiply through by `(x+5)(x^2+1)`

`(3x^2+2x-1)=A(x^2+1)+(Bx+C)(x+5)`

First solve for `A`

`x=-5`

`(3(-5)^2+2(-5)-1)=A((-5)^2+1)+(B(-5)+C)((-5)+5)`

`(3(25)-10-1)=A(25+1)+(-5B+C)(0)`

`(75-10-1)=A(26)+(-5B+C)(0)`

`64=26A`

`64/26=(26A)/26`

`32/13=A`

Now solve for `C`

Substitute in for `x` and `A`

`x=0, A=32/13`

`(3(0)^2+2(0)-1)=(32/13)((0)^2+1)+(B(0)+C)((0)+5)`

`(-1)=(32/13)(1)+(C)(5)`

`-1=32/13+5C`

`-1-32/13=5C`

`(-13)/13-32/13=5C`

`(-45)/13=5C`

`((-45)/13)/5=(5C)/5`

`((-45)/13)*1/5=C`

`((-9)/13)*1/1=C`

`(-9)/13=C`

Substitute in `A` and `C` and let `x=1` as you solve for `B`

`(3(1)^2+2(1)-1)=(32/13)((1)^2+1)+(B(1)+((-9)/13))((1)+5)`

`(3+2-1)=(32/13)(1+1)+(B-9/13)(1+5)`

`4=(32/13)(2)+(B-9/13)(6)`

`4=(64/13)+(6B-54/13)`

`4-64/13+54/13=6B`

`52/13-64/13+54/13=6B`

`106/13-64/13=6B`

`42/13=6B`

`(42/13)/6=(6B)/6`

`(42/13)*1/6=B`

`7/13=B`

Substitute in `A`, `B`, and `C`

`(3x^2+2x-1)/((x+5)(x^2+1))=(32/13)/(x+5)+((7/13)x+(-9)/13)/(x^2+1)`

`(3x^2+2x-1)/((x+5)(x^2+1))=32/(13(x+5))+(7x+(-9))/(13(x^2+1))`

Partial-Fraction Decomposition

`(3x^2+2x-1)/((x+5)(x^2+1))=32/(13(x+5))+(7x-9)/(13(x^2+1))`