#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))#
