Question

How do you evaluate `\int \frac { x ^ { 2} + 9x + 2} { ( x - 1) ( x + 1) ( x + 2) }`?

Answer 1

`int \ (x^2 + 9x + 2) / (( x - 1) ( x + 1) ( x + 2) ) \ dx = 2ln|x-1| +3ln|x+1|-4ln|x+2| + C`

Explanation:

We want to find:

`I = int \ (x^2 + 9x + 2) / (( x - 1) ( x + 1) ( x + 2) ) \ dx`

We can expand the integrand into partial fractions:

`(x^2 + 9x + 2) / ((x-1)(x+1)(x+2) ) -= A/(x-1) +B/(x+1)+C/(x+2)`
`" " = (A(x+1)(x+2) +B(x-1)(x+2)+C(x+1)(x-1)) / ((x-1)(x+1)(x+2) )`

Leading to:

`x^2 + 9x + 2 = A(x+1)(x+2) +B(x-1)(x+2)+C(x+1)(x-1)`

To find the coefficients `A`, `B` and `C` we can use substitution:

Put `x=1 \ \ \ \ \ => 1 + 9 + 2 = A(2)(3) \ \ \ \ => A =2`
Put `x=-1 => 1 - 9 + 2 = B(-2)(1) => B =3`
Put `x=-2 => 4 -18 + 2 = C(-1)(-3) => C =-4`

Hence we the partial fraction decomposition is:

`(x^2 + 9x + 2) / ((x-1)(x+1)(x+2) ) -= 2/(x-1) +3/(x+1)-4/(x+2)`

So we can write the integral as:

`I = int 2/(x-1) +3/(x+1)-4/(x+2) \ dx`

Which we can just integrate to get:

`I = 2ln|x-1| +3ln|x+1|-4ln|x+2| + C`