Question

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

Answer 1

`I=x-2/3ln(x-1)-10/3ln(x+2)+C,` OR
`I=x-2/3ln(x-1)(x+2)^5+C,` OR
`I=x-ln(x-1)^(2/3)(x+2)^(10/3)+C.`

Explanation:

Let `I=int(x^2-3x)/{(x-1)(x+2)}dx=int(x^2-3x)/(x^2+x-2)dx.`

The Degree of Poly. in `Nr.` = `2` = that of poly in `Dr.`

Hence, it is Improper Rational fun. To make it Proper, usually Long Division is performrd, but, here we proceed as under :

We write, `Nr.=x^2-3x=x^2+x-2 [i.e., = Dr]-4x+2`, so

`(x^2-3x)/(x^2+x-2)={(x^2+x-2)-(4x-2)}/(x^2+x-2)`
`=(x^2+x-2)/(x^2+x-2)-(4x-2)/(x^2+x-2)=1-(4x-2)/(x^2+x-2)`
`=1-(4x-2)/{(x-1)(x+2)}.`

Hence, `I=int[1-(4x-2)/{(x-1)(x+2)}]dx=int1dx-I_1=x-I_1,`

where `I_1=int(4x-2)/{(x-1)(x+2)}dx`

To evaluate `I_1`, we have to split `(4x-2)/{(x-1)(x+2)}......(1)` using Partial Fractions as `A/(x-1)+B/(x+2)={A(x+2)+B(x-1)}/{(x-1)(x+2)}.....(2); A,B in RR.`

Since `(1) and (2)` are equal, we get `: A(x+2)+B(x-1)=4x-2, AAx.`

In particular, `x=1 rArr 3A=2 rArr A=2/3` and `x=-2 rArr B=10/3.`

Accordingly, `I_1=int(2/3)/(x-1)dx+int(10/3)/(x+2)dx`
`=2/3ln|x-1|+10/3ln|x+2|.` Finally, we altogether get,

`I=x-2/3ln(x-1)-10/3ln(x+2)+C`, or, using the Rules of Log,

`I=x-2/3ln(x-1)(x+2)^5+C=x-ln(x-1)^(2/3)(x+2)^(10/3)+C`