Evaluate the integral integrate of ((3)/(x^2-x-2))dx

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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:

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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` .

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