Evaluate `int((3x+1))/((2x^(2)-2x+3))dx`.

Explanation

Let `(3x+1)=A*(d)/(dx)(2x^(2)-2x+3)+B`.Then , `(3x+1)=A(4x-2)+B` " " (i)
Comparing the coefficients of like powers of x we get
`(4A=3andB-2A=1)rArr(A=(3)/(4)andB=(5)/(2))`.
`thereforeint((3x+1))/((2x^(2)-2x+3))dx=int(A*(4x-2)+B)/((2x^(2)-2x+3))`
`=int((3)/(4)*(4x-2)+(5)/(2))/((2x^(2)-2x+3))dx=(3)/(4)*int((4x-2))/((2x^(2)-2x+3))dx+(5)/(2)int(dx)/(2(x^(2)-x+(3)/(2)))`
`=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*int(dx)/({(x^(2)-x+(1)/(4))+(5)/(4)})`
`=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*int(dx)/({(x-(1)/(2))^(2)+((sqrt(5))/(2))^(2)})`
`=(3)/(4)log|2x^(2)-2x+3|+(5)/(4)*(1)/(((sqrt(5))/(2)))"tan"^(-1){((x-(1)/(2)))/(((sqrt(5))/(2)))}+C`
`=(3)/(4)log|2x^(2)-2x+3|+(sqrt(5))/(2)tan^(-1)((2x-1)/(sqrt(5)))+C`.