Question

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

Answer 1

`(2x^2+5)/(x^2+1)^2=3/(x^2+1)^2+2/(x^2+1)`

Explanation:

Begin with the set up of the variables A, B, C, D

`(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx+D)/(x^2+1)`

the LCD(Least common Denominator) is `=(x^2+1)^2`

so that the right side of the equation becomes

`(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+((Cx+D)(x^2+1))/(x^2+1)^2`

expand

`(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx^3+Cx+Dx^2+D)/(x^2+1)^2`

combine

`(2x^2+5)/(x^2+1)^2=(Ax+B+Cx^3+Cx+Dx^2+D)/(x^2+1)^2`

rearrange

`(0*x^3+2x^2+0*x+5*x^0)/(x^2+1)^2=(Cx^3+Dx^2+(A+C)x+(B+D)x^0)/(x^2+1)^2`

the equations can now be found by equating the numerical coefficients:

`C=0`
`D=2`
`A+C=0`
`B+D=5`

Solving for A, B, C, D:

`A=0` and `B=3` and `C=0` and `D=2`

so that the partial fractions are

`(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx+D)/(x^2+1)=(0*x+3)/(x^2+1)^2+(0*x+2)/(x^2+1)`

and

`(2x^2+5)/(x^2+1)^2=3/(x^2+1)^2+2/(x^2+1)`

have a nice day! from the Philippines..