How do you integrate #(x^3 - 2) / (x^4 - 1)# using partial fractions?
1 Answer
Explanation:
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
Substituting
Substituting these values into
Furthermore, since
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
#=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#