Question

How do you integrate `x / (x² - x - 2)` using partial fractions?

Answer 1

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

Explanation:

First, separate the fraction using partial fraction separation:

`x/(x^2-x-2) = x/((x-2)(x+1))`

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

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

Set `x=2` to solve for A:

`2 = A(2+1) + B(2-2)`

`2 = 3A`

`2/3 = A`

Set `x=-1` to solve for B:

`-1 = A(-1+1) + B(-1-2)`

`-1 = -3B`

`1/3 = B`

Therefore, our fraction separates into:

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

Finally, integrate:

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

`= 2/3ln|x-2| + 1/3ln|x+1| + C`

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

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

Final Answer