f(x)=根号3sin2x+2cos平方x+1
=√3sin2x+cos2x+2
=2sin(2x+π/6)+2
所以
当2kπ-π/2<2x+π/6<2kπ+π/2时,单调递增,此时
2kπ-2π/3<2x<2kπ+π/3
即
kπ-π/3x∈（kπ-π/3,kπ+π/6）为单调增区间.