Question

How do you integrate `int (x+4) /((x+1)(x - 2)²)` using partial fractions?

Answer 1

Solution is

`1/3*ln|x+1| - 1/3*ln|x-2| - 2/(x-2)`

Explanation:

This could become quite lengthy, just warning you.

`(x+4)/((x+1)(x-2)^2) = A/(x+1) + B/(x-2) + C/((x-2)^2)`

where A,B and C are arbitrary constants.

Multiplying appropriately to get common denominators we get:

`x+4 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)`

There are a number of methods which can be used to determine the values of the constants, we shall set the value of x such that certain brackets evaluate to zero:

`x = 2`

`therefore 2 + 4 = 3*C implies C = 2`

`x = -1`

`therefore -1 + 4 = 9*A implies A = 1/3`

Now we know A and C, try at x = 0:

`0 + 4 = 1/3*(-2)^2 - 2*B + 2`

`4 = 4/3 + 2 - 2*B implies B = -1/3`

So the integral has become:

`1/3*int (dx)/(x+1) - 1/3*int (dx)/(x-2) + 2*int (dx)/((x-2)^2)`

For the first part, let `u = x + 1 implies du = dx`
for the second and third, let `v = x - 2 implies dv = dx`

`1/3*int (du)/(u) - 1/3*int (dv)/(v) + 2*int (dv)/(v^2)`

`= 1/3*ln|x+1| - 1/3*ln|x-2| - 2/(x-2)`