How do you evaluate the integral #int 1/(x^3-x)dx#?

1 Answer
Mar 6, 2017

#-lnabsx+(1/2)lnabs(x-1)+(1/2)lnabs(x+1)+C#

Explanation:

This can be solved by using the method of partial fractions.

#int1/(x^3-x)dx#

First factor the denominator completely.

#int1/(x(x^2-1))dx=int1/(x(x-1)(x+1))dx#

Decompose the integrand.

#1/(x(x+1)(x-1))=A/x+B/(x-1)+C/(x+1)#

Get the lowest common denominator between all of the terms of the decomposition.

#1/(x(x+1)(x-1))=(A(x-1)(x+1))/(x(x-1)(x+1))+(Bx(x+1))/(x(x-1)(x+1))+(Cx(x-1))/(x(x-1)(x+1))#

Now that all terms have a common denominator, focus on the numerator to solve for the value of A, B, and C.

#1=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)#

For #x=0#,
#1=A(-1)(1)#
#1=-A#
#-1=A#

For #x=1#,
#1=B(1)(2)#
#1=2B#
#1/2=B#

For #x=-1#,
#1=C(-1)(-2)#
#1=2C#
#1/2=C#

Substitute the decomposition in to the integral.

#int (-1)/x+(1/2)/(x-1)+(1/2)/(x+1)dx#

Break the integral into smaller integrals for each term

#int (-1)/xdx+int(1/2)/(x-1)dx+int(1/2)/(x+1)dx#

Bring out the constants

#-int1/xdx+1/2int1/(x-1)dx+1/2int1/(x+1)dx#

Integrate each term

#-lnabsx+(1/2)lnabs(x-1)+(1/2)lnabs(x+1)+C#