Question

How do you integrate `int (x^3 + 2x - 1) / (2x^2 - 3x - 2)` using partial fractions?

Answer 1

`x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.`

Explanation:

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

`" "x^3+0x^2+2x-1" "`|`" "2x^2-3x-2`
`-x^3+3/2x^2+x" "`|____
_____#" "1/2x+3/4#
`" "3/2x^2+3x-1`
`" "-3/2x^2+9/4x+3/2`
`" "` _______
`" "21/4x+1/2=1/4(21x+2)`

So the expression becomes
`=int(x/2+3/4)dx+1/4int (21x+2)/(2x^2-3x-2)dx`

Let's deal with the last part
Finding the zeros of the denominator
`2x^2-3x-2=0`

`Delta=9+16=25` => `sqrt(Delta)=5`
`x=(3+-5)/4` => `x_1=-1/2`; `x_2=2`

Then we can break the second integrand in this way:
`(21x+2)/(2x^2-3x-2)=A/(x+1/2)+B/(x-2)`
To find `A` and `B` let's make `x=0 and -1`

`-1=2A-B/2`
`-19/3=-2A-B/3`
Summing these 2 expressions we get
`-22/3=-5/6B` => `B=44/5`
`-> A=(-1+B/2)/2=(-1+22/5)/2` => `A=17/10`

So the main expression becomes
`=x^2/4+(3x)/4+(1/4)(17/10int dx/(x+1/2)+44/5int dx/(x-2))`
`=x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.`