Question

How do you express `(3x )/ (x^2 * (x^2+1) )` in partial fractions?

Answer 1

`(3x)/(x^2*(x^2+1))=3/x-(3x)/(x^2+1)`

Explanation:

`(3x)/(x^2*(x^2+1))=3/(x*(x^2+1)` and let its partial fractions be

`3/(x*(x^2+1))hArrA/x+(Bx+C)/(x^2+1)` or

`3/(x*(x^2+1))hArr(A(x^2+1)+x(Bx+C))/(x*(x^2+1))` or

`3/(x*(x^2+1))hArr(x^2(A+B)+Cx+A)/(x*(x^2+1))` or

`A+B=0`, `C=0` and `A=3`. Thus `B=-3`

Hence `(3x)/(x^2*(x^2+1))=3/x-(3x)/(x^2+1)`