Question

How do you evaluate the integral `int dx/((2x-1)(x+2))`?

Answer 1

Use partial fractions. See below.

`int1/((2x-1)(x+2))dx=1/5ln(|2x-1|)-1/5ln(|x+2|)+C`

Explanation:

The denominator is already factored.

`=>1/((2x-1)(x+2))=A/(2x-1)+B/(x+2)`

Multiply through by the denominator of the left-hand side:

`(2x-1)(x+2)[1/cancel((2x-1)(x+2))=A/(cancel(2x-1))+B/cancel(x+2)]`

`=>1=A(x+2)+B(2x-1)`

We need to solve for `A` and `B`. One way to do this is to pick values for `x` which will cancel each variable.

`x=-2`

`1=B(-4-1)`

`1=-5B`

`B=-1/5`

`x=1/2`

`1=A(1/2+2)`

`1=A(5/2)`

`A=2/5`

We put these values back into our partial fractions and replace this as the integrand.

`=>int(2/5)/(2x-1)+(-1/5)/(x+2)dx`

Technically you should use a substitution before integrating. Split up the integral.

`=>2/5int1/(2x-1)dx-1/5int1/(x+2)dx`

For the first integral, `u=2x-1, du=2dx=> 1/2du=dx`.

For the second integral, `z=x+2, dz=dx`.

`=>1/5int1/udu-1/5int1/zdz`

Integrate.

`=>1/5ln|u|-1/5ln|z|+C`

Substitute back in.

`=>1/5ln|2x-1|-1/5ln|x+2|+C`

Note: the absolute value signs account for the domain of the natural log function (`x>0`).

By rules of logarithms, this is equivalent to

`1/5ln(|(2x-1)/(x+2)|)+C`