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

We see that:

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

Split it up into its partial fractions:

`(x^3-2)/((x^2+1)(x+1)(x-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)`

Multiplying through, we see that:

`x^3-2=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)`

Continue:

`x^3-2=(Ax+B)(x^2-1)+C(x^3-x^2+x-1)+D(x^3+x^2+x+1)`

`x^3-2=Ax^3-Ax+Bx^2-B+Cx^3-Cx^2+Cx-C+Dx^3+Dx^2+Dx+D`

`x^3-2=x^3(A+C+D)+x^2(B-C+D)+x(-A+C+D)+(-B-C+D)`

Equating the coefficients on either side of the equation:

`{(1=A+C+D),(0=B-C+D),(0=-A+C+D),(-2=-B-C+D):}`

Adding the first and third equations, we see that `1=2C+2D`. Adding the second and fourth equations, we see that `-2=-2C+2D`. Adding these two equations, this yields `-1=4D`, so `D=-1/4`.

Substituting `D=-1/4` into `1=2C+2D`, we see that `C=3/4`.

Substituting these values into `1=A+C+D`, this yields that `A=1/2`.

Furthermore, since `0=B-C+D`, we see that `B=1`.

Thus:

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

So:

`int(x^3-2)/(x^4-1)dx=1/2int(x+2)/(x^2+1)dx+3/4int1/(x+1)dx-1/4int1/(x-1)dx`

Split up the first integral. The last two integrals can be integrated simply:

`=1/2intx/(x^2+1)dx+1/2int2/(x^2+1)dx+3/4lnabs(x+1)-1/4lnabs(x-1)`

Rearranging the first term to set up for a natural logarithm substitution, since `2x` is the derivative of `x^2+1`:

`=1/4int(2x)/(x^2+1)dx+int1/(x^2+1)dx+3/4lnabs(x+1)-1/4lnabs(x-1)`

The second integral is a common integral as well:

`=1/4ln(x^2+1)+arctan(x)+3/4lnabs(x+1)-1/4lnabs(x-1)+C`