分母先配方,得x^2-x+1=(x-1/2)^2+3/4
然后先求不定積分,按不定積分換元,x-1/2=√3/2*tanu,則u=2/√3*arctan(x-1/2),
由x-1/2=√3/2*tanu 可得：sinu=(x-1/2)/(x^2-x+1)
dx=√3/2*(secu)^2du
∫ 1/[(x-1/2)^2+3/4]^(3/2) dx
=∫ 1/[3/4*(tanu)^2+3/4]^(3/2) *√3/2*(secu)^2du
=∫1/[(3/4)^(3/2)*(sec)^3] *√3/2*(secu)^2 du
=4/3 *∫cosu du=4/3sinu+C
=4/3*(x-1/2)/(x^2-x+1)+C
將x用1,0代入相減得：2/3-(-2/3)=4/3