求不定积分:1/(x+sqrt(x^2-x+1))

1/(x+sqrt(x^2-x+1))=[sqrt(x^2-x+1)-x](1-x)
sqrt(x^2-x+1)(1-x)dx=-sqrt(t^2-t+1)tdt t=1-x
-sqrt(t^2-t+1)t=12sqrt(t^2-t+1)-(t-12)sqrt(t^2-t+1)
-1tsqrt(t^2-t+1)
u=1t -1tsqrt(t^2-t+1)dt=sgnusqrt(u^2-u+1)du
原式=∫12sqrt(t^2-t+1)-(t-12)sqrt(t^2-t+1)dt+∫sgnusqrt(u^2-u+1)du
+x+In|x-1|+C
=(12)In[t-12+sqrt(t^2-t+1)]-sqrt(t^2-t+1)+In|1t-12+sqrt(t^2-t+1)t|
+x+In|x-1|+C
=(12)In[sqrt(x^2-x+1)-x+12]+In[sqrt(x^2-x+1)+(x+1)2]+x-sqrt(x^2-x+1)
+C