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

Explanation

`"Let I"=int (2x^(3)+3x^(2)-3)/(2x^(2)-x-1)dx`
`therefore I=int [(x+2)+(3x-1)/(2x^(2)-x-1)]dx`
`=int xdx+2int 1dx+int (3x-1)/(2x^(2)-x-1)dx`
`"Let " (3x-1)/(2x^(2)-x-1)=(3x-1)/((x-1)(2x+1))=(A)/(x-1)+(B)/(2x+1)`
`therefore 3x-1=A(2x+1)+B(x-1)`
Put x-1=0 i.e, x=1 we get
`2=A(3)+B(0)" "therefore A=(2)/(3)`
Put 2x+1=0, i.e, `x=-(1)/(2)`, we get
`-(3)/(2)-1=A(0)+B(-(1)/(2)-1)" "therefore -(5)/(2)=-(3)/(2)B" "therefore B=(5)/(3)`
`therefore (3x-1)/(2x^(2)-x-1)=((2//3))/(x-1)+((5//3))/(2x+1)`
`therefore I=int xdx+2int 1dx+int [((2//3))/(x-1)+((5//3))/(2x+1)]dx`
`=int dx+2int 1dx+(2)/(3)int (1)/(x-1)dx+(5)/(3)int (1)/(2x+1)dx`
`=(x^(2))/(2)+2x+(2)/(3)log |x-1|+(5)/(3).(log|2x+1|)/(2)+c`
`=(x^(2))/(2)+2x+(2)/(3)log |x-1|+(5)/(6)log|2x+1|+c`