Question

How do you integrate `int(x^2) / ( x^2 + 3x +2) dx` using partial fractions?

Answer 1

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

Explanation:

Original
`int x^2/(x^2+3x+2)dx " " " " " " (1)`

Step 1: Rewrite the expression as a proper fraction (using long division)

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

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

Step 2: Factor the denominator `(3)`

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

Set up the partial fraction decomposition
`(3x+2)/(x^2 + 3x +2) =`

`=(3x+2)/((x+1)(x+2)) = color(red)A/(x+2) + color(red)B/(x+1) " " " " " " (4)`

Multiply LCD (least common denominator)

`(3x+2) = A(x+1) + B(x+2)`

Let `x= -1`
`(3*-1+2)= A(-1+1)+B(-1+2)`
`color(red)(-1= B)`

Let `x= -2`
`3*-2+2= A(-2+1)+B(-2+2)`
`-4 = -A`
`color(red)(4=A)`

Rewrite `(3)`
`color(red)4/(x+2) - color(red) (1)/(x+1)`

Rewrite (2)

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

`=int 1*dx -int 4/(x+2) dx +int 1/(x+1) dx`

`color(blue)( =x -4 ln|x+2| -ln|x+1| +C) " " " " " (5)`

I hope this help.