Question

How do you express `6/(x^2-25)^2` in partial fractions?

Answer 1

The answer is
`6/(x^2-25)^2=3/(50(x+5)^2)+3/(250(x+5))+3/(50(x-5)^2)-3/(250(x-5))`

Explanation:

we start by factorising the denominator
`(x^2-25)^2=((x+5)(x-5))^2`
So we have`=6(1/((x+5)^2(x-5)^2))`
Then we can do the decomposition in partial fractions
`1/((x+5)^2(x-5)^2)=A/((x+5)^2)+B/(x+5)+C/((x-5)^2)+D/(x-5)`

`=((A(x-5)^2+B(x+5)(x-5)^2+C(x-5)^2+D(x-5)(x+5)^2))/(((x+5)^2(x-5)^2))`

So `1=(A(x-5)^2+B(x+5)(x-5)^2+C(x+5)^2+D(x-5)(x+5)^2))`
Let `x=5` we get `1=100C` `=>` `C=1/100`
Let `x=-5` then `1=100A` `=>` `A=1/100`
Let `x=0` then `1=25A+125B+25C-125D`
`1=25/100+25/100+125(B-D)`
`B-D=1/250`
Coefficients of `x^3` `=>` `0=B+D` `=>` `B=-D`
`B=1/500` and `D=-1/500`
So the final result is `=6/(100(x+5)^2)+6/(500(x+5))+6/(100(x-5)^2)-6/(500(x-5))`