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#