To integrate this, we need to separate the function into simpler functions using partial fractions, then integrate each function separately.

`int 3/{x^2-x-2}dx` factor denominator

`=int 3/{(x-2)(x+1)}dx` split into partial fractions

this means that `3/{(x-2)(x+1)}=A/{x-2}+B/{x+1}` . Multiplying out and comparing on the constant terms and the linear terms, we see that:

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

To integrate this, we need to separate the function into simpler functions using partial fractions, then integrate each function separately.

`int 3/{x^2-x-2}dx` factor denominator

`=int 3/{(x-2)(x+1)}dx` split into partial fractions

this means that `3/{(x-2)(x+1)}=A/{x-2}+B/{x+1}` . Multiplying out and comparing on the constant terms and the linear terms, we see that:

`3=A(x+1)+B(x-2)` which gives the two equations:

`3=A-2B` equation (1)

`0=A+B` equation (2)

Subtract (1) from (2)

`3B=3` so B=1 and A=-1

Now the integral becomes

`-int 1/{x-2} dx + int1/{x+1}dx` integrate each term

`=-ln(x-2)+ln(x+1)+C` where C is the constant of integration

`=ln({x+1}/{x-2})+C`

**The integral evaluates to `ln({x+1}/{x-2})+C` .**