Question

怎么积分的?另外这种类型有什么思路?

Answer 1

x^2+x+1

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

所以原式=∫d(x+1/2)/[(x+1/2)^2+(√3/2)^2]

=1/(√3/2)*arctan[(x+1/2)/(√3/2)]+C

=2/√3*arctan[(2x+1)/√3]+C