Question

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

Answer 1

`-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`