Evaluate

(x²-x+1) / (2x²+x+2) = (2x²-2x+2) /[2(2x²+x+2)]     ...........multiply and divide by 2

= (2x²+x+2 - 3x) /[2(2x²+x+2)]

=(1/2)[1 - 3x/(2x²+x+2)]

= (1/2)[1 - (3/4)(4x+1-1)/(2x²+x+2)]                  ..........3x = (3/4)(4x+1-1)

= (1/2)[1 - (3/4)(4x+1)/(2x²+x+2) +(3/4)/(2x²+x+2)]

(x²-x+1) dx/ (2x²+x+2) =

(1/2)[1 - (3/4)(4x+1)/(2x²+x+2) +(3/4)/(2x²+x+2)]dx =

(1/2){dx - (3/4)(4x+1)dx/(2x²+x+2) + (3/4)dx/(2x²+x+2)} = 

(1/2){x - (3/4) ln (2x²+x+2) + (3/4)dx/(2x+1/22)² + ((15/8))²}

= (1/2){x - (3/4) ln (2x²+x+2) + (3/4)((8/15))(2)tan^-1((2x + 1/22)/(15/8))} + c

= (1/2){x - (3/4) ln (2x²+x+2) + (3/5) tan^-1((4x + 1 )/(15))} + c

Regards,

Team,

TopperLearning.