∫[(x+1)/x²-x+1)]dx=?

x^2-x+1=x^2-x+1/4+3/4
=(x-1/2)^2+3/4

设t=x-1/2

∫[(x+1)/(x²-x+1)]dx
=∫[(t+3/2)/(t²+3/4)]dt
=∫t/(t²+3/4)dt+3/2∫1/(t²+3/4)]dt
=∫1/(t²+3/4)d(t^2+3/4)+3/2∫1/(t²+3/4)dt
=ln/(t^2+3/4)/+3/2∫1/(t²+3/4)dt

对于第二项
设t=根号3/2*tan(u)
dt=根号3/2*(sec(u))^2du

后面就简单了

ln|x|-1/x-x^2/2+x+c

∫[(x+1)/x²-x+1)]dx=∫(1/x)dx+∫(1/x²)dx-∫xdx+∫dx
=ln1x1+(1/x)-(x²/2)+x+c