Question

How do you integrate `int(x^3+1)/((x^2-4)(x^2+1)) dx` using partial fractions?

Answer 1

`int(x^3+1)/((x^2-4)(x^2+1))`

= `9/20ln|x-2|+7/20ln|x+2|+1/10ln|x^2+1|+1/5tan^(-1)x`

Explanation:

As `(x^3+1)/((x^2-4)(x^2+1))=(x^3+1)/((x-2)(x+2)(x^2+1))`

Let `(x^3+1)/((x^2-4)(x^2+1))=A/(x-2)+B/(x+2)+(Cx+D)/(x^2+1)`

or `(x^3+1)=A(x+2)(x^2+1)+B(x-2)(x^2+1)+(Cx+D)(x+2)(x-2)`

when `x=2` we have `9=20A` i.e. `A=9/20`

when `x=-2` we have `-7=-20B` i.e. `B=7/20`

Comparing coefficients of `x^3` and constant term we get

`1=A+B+C` i.e. `C=1-A-B=1-9/20-7/20=4/20=1/5`

and `1=2A-2B-4D` i.e. `D=-(1-18/20+14/20)/4=1/5`

Hence `(x^3+1)/((x^2-4)(x^2+1))=9/(20(x-2))+7/(20(x+2))+(x+1)/(5(x^2+1))` and

`int(x^3+1)/((x^2-4)(x^2+1))dx`

= `int9/(20(x-2))dx+int7/(20(x+2))dx+int(x+1)/(5(x^2+1))dx`

= `9/20ln|x-2|+7/20ln|x+2|+1/5(1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx)`

= `9/20ln|x-2|+7/20ln|x+2|+1/10ln|x^2+1|+1/5tan^(-1)x`