Integrate the following w.r.t.x:
`(x^(2)+2)/((x-1)(x+2)(x+3))`

Explanation

`"Let I"=int (x^(2)+2)/((x-1)(x+2)(x+3))dx`
`"Let "(x^(2)+2)/((x-1)(x+2)(x+3))=(A)/(x-1)+(B)/(x+2)+(C)/(x+3)`
`therefore x^(2)+=A(x+2)(x+3)+B(x-1)(x+3)+C(x-1)(x+2)`
`"Put "x-1=0,i.e. x=1"we get"`
`1+2=A(3)(4)+B(0)(4)+C(0)(3)`
`4+2=A(3)(4)+B(0)(4)+C(0)(3)`
`therefore 3=12A " "A=(1)/(4)`
`"Put x+2=0, i.e, x=-2, we get "`
`4+2=A(0)(1)+B(-1)(1)+C(-3)(0)`
`therefore 6=-3B" "therefore B=2`
`"Put x+3=0, i.e, x=-3 we get"`
`9+2=A(-1)(0)+B(-4)(0)+C(-4)(-1)`
`therefore 11=4C" "therefore C=(11)/(4)`
`therefore (x^(2)+2)/((x-1)(x+2)(x+3))=((1//4))/(x-1)+((-2))/(x+2)+((11//4))/(x+3)`
`therefore I=int [((1//4))/(x-1)+((-2))/(x+2)+((11//4))/(x+3)]dx`
`=(1)/(4) int (1)/(x-1)dx-2 int (1)/(x+2)dx+(11)/(4)int (1)/(x+3)dx`
`=(1)/(4)log|x-1|-2log|x+2|+(11)/(4)log|x+3|+c`