int(-2x^3-x^2+x+2)/(2x^2-x+3)dx
=int(-2x^3+x^2-3x)/(2x^2-x+3)+(-2x^2+4x+2)/(2x^2-x+3)dx
(separate numerator)
=int-x+(-2x^2+x-3)/(2x^2-x+3)+(3x+5)/(2x^2-x+3)dx (factor out 2x^2-x+3 from 1st fraction and separate second fraction so the numerator will divide evenly by the denominator)
=-1/2x^2+int-1+(3x-3/4)/(2x^2-x+3)+(23/4)/(2x^2-x+3)dx (integrate -x with reverse power rule, factor out 2x^2-x+3 from 1st fraction, and separate second fraction. you don't need to write +C yet because another +C will be created from integrating the rest of the expression)
=-1/2x^2-x+int3/4((4x-1)/(2x^2-x+3))+(23/4)/((sqrt(2)x-sqrt(2)/4)^2+23/8)dx (integrate -1 and rewrite the fractions so it will be easier to integrate to ln(f(x)) and arctan(f(x)))
=-1/2x^2-x+3/4ln|2x^2-x+3|+int(23/4)/((sqrt(2)x-sqrt(2)/4)^2+23/8)dx (integrate the first fraction using int(f'(x))/f(x)dx=ln(f(x))
=-1/2x^2-x+3/4ln(2x^2-x+3)+23/4*1/sqrt(2)intsqrt(2)/((sqrt(2)x-sqrt(2)/4)^2+(sqrt(23/8))^2)dx (change coefficients to prepare for integration to arctan(f(x)) and change absolute value to parentheses because 2x^2-x+3 is always positive)#
=-1/2x^2-x+3/4ln(2x^2-x+3)+(23/4)/(sqrt(2*23/8))arctan((sqrt(2)x-sqrt(2)/4)/(sqrt(23/8)))+C (integrate the fraction using int(f'(x))/((f(x))^2+a^2)dx=1/aarctan(f(x)/a)+C where a is a constant)
=-1/2x^2-x+3/4ln(2x^2-x+3)+(23/4)/(sqrt(23/4))arctan((sqrt(2*8)x-sqrt(2*8)/4)/(sqrt(23))) (simplify)
=-1/2x^2-x+3/4ln(2x^2-x+3)+sqrt(23/4)arctan((4x-1)/sqrt(23))
=-1/2x^2-x+3/4ln(2x^2-x+3)+sqrt(23)/2arctan((4x-1)/sqrt(23))
