Question

How do you integrate `int(x)/((x+3)(x+6)(x+2))` using partial fractions?

Answer 1

`int x/((x+3)(x+6)(x+2)) dx= ln|x+3| - 1/2ln|x+6| - 1/2ln|x+2| + C`

Explanation:

The Partial Fraction decomposition of the integrand will be of the form:

`x/((x+3)(x+6)(x+2)) -= A/(x+3) + B/(x+6) + C/(x+2)`
`:. x/((x+3)(x+6)(x+2)) = (A(x+6)(x+2) + B(x+3)(x+2) + C(x+3)(x+6))/((x+3)(x+6)(x+2))`
`:. x = A(x+6)(x+2) + B(x+3)(x+2) + C(x+3)(x+6)`

Put `{ (x=-3, =>-3=A(3)(-1), => A=1), (x=-6,=>-6=B(-3)(-4),=>B=-1/2),(x=-2,=>-2=C(1)(4),=>C=-1/2) :}`

Hence,

`x/((x+3)(x+6)(x+2)) -= 1/(x+3) - 1/(2(x+6)) - 1/(2(x+2))`

So,

`int x/((x+3)(x+6)(x+2)) dx= int 1/(x+3) - 1/(2(x+6)) - 1/(2(x+2)) dx`
`:. int x/((x+3)(x+6)(x+2)) dx= ln|x+3| - 1/2ln|x+6| - 1/2ln|x+2| + C`