∫_1^√3 1/[x√(1+x)] dx x=tany,dx=secydy =∫_(π/4)^(π/3) 1/(tanysecy) * secydy =∫_(π/4)^(π/3) 1/cosy*cosy/siny dy =∫_(π/4)^(π/3) cscycotydy =-cscy |_(π/4)^(π/3) = -[1/sin(π/3) - 1/sin(π/4)] =√2 - 2/√3
记得采纳啊
∫_1^√3 1/[x√(1+x)] dx x=tany,dx=secydy =∫_(π/4)^(π/3) 1/(tanysecy) * secydy =∫_(π/4)^(π/3) 1/cosy*cosy/siny dy =∫_(π/4)^(π/3) cscycotydy =-cscy |_(π/4)^(π/3) = -[1/sin(π/3) - 1/sin(π/4)] =√2 - 2/√3
记得采纳啊