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

Explanation

`"Let I" =int (2x+3)/(2x^(2)+3x-1)dx`
`"Let "2x+3=A[(d)/(dx)(2x^(2)+3x-1)]+B`
`=A(4x+3)+B`
`therefore 2x+3=4Ax+(3A+B)`
Comparing the coefficient of x and constant on both sides, we get 4A=2 and 3A+B=3
`therefore A=(1)/(2) and 3((1)/(2))+B=3 " "therefore B=(3)/(2)`
`therefore 2x+3=(1)/(2)(4x+3)+(3)/(2)`
`therefore I=int ((1)/(2)(4x+3)+(3)/(2))/(2x^(2)+3x-1)dx`
`therefore =(1)/(2)int (4x+3)/(2x^(2)+3x-1)dx+(3)/(2) int (1)/(2x^(3)+3x-1)dx`
`=(1)/(2)I_(1)+(3)/(2)I_(2)`
`I_(1)` is of the type `int (f'(x))/(f(x))dx=log|f(x)|`
`therefore I_(1)=log|2x^(2)+3x-1|`
`I_(2)=int (1)/(2x^(2)+3x-1)dx=(1)/(2)int (1)/(x^(2)+(3)/(2)x-(1)/(2))dx`
`=(1)/(2)int (1)/((x^(2)+(3)/(2)x+(9)/(16))-(9)/(16)-(1)/(2))dx`
`=(1)/(2)int (1)/((x+(3)/(4))^(2)-((sqrt17)/(4))^(2))dx`
`=(1)/(2)xx(1)/(2xx(sqrt17)/(4))log |(x+(3)/(4)-(sqrt17)/(4))/(x+(3)/(4)+(sqrt17)/(4))|`
`=(1)/(17)log |(4x+3-sqrt17)/(4x+3+sqrt17)|`
`therefore I=(1)/(2)log|2x^(2)+3x-1|+(3)/(2sqrt17)log|(4x+3-sqrt17)/(4x+3+sqrt17)+c`