Question

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

Answer 1

`-2/3ln|x+1|+1/3ln|x^2-x+1|+2/sqrt3tan^-1((2x-1)/sqrt3)+C`

Explanation:

The denominator factorises as `(x+1)(x^2-x+1)`. So the integrand is
`(-2/3)/(x+1)+(Ax+B)/(x^2-x+1)` where the `2/3` comes from the cover-up rule: `((2)(-1))/((-1)^2-(-1)+1)` and `A` and `B` are to be found by equating coefficients of suitable powers of `x` or otherwise:
`2x-=(-2/3)(x^2-x+1)+(Ax+B)(x+1)` which, upon re-grouping in powers of `x` gives:
`0x^2+2x+0-=(-2/3+A)x^2+(2/3+A+B)x-2/3+B`.
So `A=2/3`, `B=2/3`

So the integral is:
`int 2/3(-1/(x+1)+1/2(2x-1+3)/(x^2-x+1))dx`
`=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+int1/((x-1/2)^2+3/4)dx`
`=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+(2/sqrt3)tan^-1((x-1/2)/(sqrt3/2))+C`
`=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+(2/sqrt3)tan^-1((2x-1)/sqrt3)+C`

Notice the trick of splitting the numerator `x+1` into the multiple of the derivative of the denominator plus a correction constant (here `+3`) in anticipation of using the rule about integrating `int(f'(x))/f(x)dx=ln|f(x)|+C`