Question

How do you find `int 8/ ( x^2 (x+3) )dx` using partial fractions?

Answer 1

Apply partial fractions to find
`int8/(x^2(x+3))dx = 8/9(ln|(x+3)/x|-3/x) + C`

Explanation:

In partial fraction decomposition, we consider denominators which could combine to create the original denominator as the common denominator, and then solve for the numerators.

As the factors of `x^2(x+3)` are `x`, `x^2`, and `x+3` we proceed as follows:

`8/(x^2(x+3)) = A/x + B/x^2 + C/(x+3)`

Now, multiplying through by `x^2(x+3)` we obtain

`8 = Ax(x+3) + B(x+3) + Cx^2`

`= (A+C)x^2 + (3A+B)x + (3B)`

Now, we equate corresponding coefficients to obtain the system
`{(A+C = 0), (3A+B = 0), (3B = 8):}`

Solving the system, we obtain
`{(A = -8/9), (B = 8/3), (C = 8/9):}`

Then we substitute back into our original equation to obtain

`8/(x^2(x+3)) = (-8/9)/x + (8/3)/x^2 + (8/9)/(x+3)`

`=> 8/(x^2(x+3)) = -8/(9x) + 8/(3x^2) + 8/(9(x+3))`

Now we proceed to integrate.

`int8/(x^2(x+3))dx = int(-8/(9x) + 8/(3x^2) + 8/(9(x+3)))dx`

`= int-8/(9x)dx + int8/(3x^2)dx + int8/(9(x+3))dx`

`= -8/9int1/xdx + 8/3int1/x^2dx + 8/9int1/(x+3)dx`

`=-8/9ln|x| + 8/3(-1/x) + 8/9ln|x+3| + C`

`= 8/9(ln|(x+3)/x|-3/x) + C`