Question

How do you evaluate the integral `int (5x+1)/((2x+1)(x-1))dx`?

Answer 1

You separate the integrand as a sum of first order rational functions using the development in partial fractions

Explanation:

Pose:

`(5x+1)/((2x+1)(x-1)) = A/(2x+1) + B/(x-1)`

Developing the second part:

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

Equate the coefficient of the same order of the numerators and get:

`A+2B =5`
`A-B=-1`

Solving this system you get:

`A=1, B=2`

so:

`int(5x+1)/((2x+1)(x-1))dx = int(1/(2x+1)+2/(x-1))dx =`
`= int dx/(2x+1)+2 int dx/(x-1) = 1/2log(2x+1)+2log(x-1) = log((x-1)^2sqrt(2x+1))`