Question

How do you integrate `int ( x+ 2 ) / (x(x-3)(x+2)) dx` using partial fractions?

Answer 1

`int (x+2)/(x(x-3)(x+2)) dx = (ln abs (x-3) - ln abs (x))/3 +C`

Explanation:

As a first step we can simplify the integrand function:

`cancel(x+2)/(x(x-3)cancel((x+2))) = 1/(x(x-3))`

Now write the integrand function as a sum of partial fraction with parametric numerators:

`1/(x(x-3)) = A/x+B/(x-3)`

Develop the sum at the second member:

`1/(x(x-3)) = (A(x-3)+Bx)/(x(x-3))`

`1/(x(x-3)) = ((A+B)x -3A)/(x(x-3))`

As the denominators are equal, the equation is satisfied if the numerators are equal:

`1 = (A+B)x -3A`

We can then equate the coefficient of the same degree in `x` to have a system of equations that lets us determine the parameters:

`{(A+B=0), (-3A=1):}`

`{(A=-1/3), (B=+1/3):}`

So:

`1/(x(x-3)) = 1/3(1/(x-3)-1/x)`

We can now integrate:

`int (x+2)/(x(x-3)(x+2)) dx = int (dx)/(x(x-3)) = 1/3 int (dx)/(x-3) -1/3 int (dx)/x = (ln abs (x-3) - ln abs (x))/3 +C`