Question

这个式子如何积分

其中 R为定值，那么这个式子如何积分？？

Answer 1

其不定积分
=1/R^2∫1/[(x/R)^2+1]^3/2 d(x/R)
令x/R=tant, -π/2<t<π/2
原式化为：
∫sec²tdt/sec³t
=∫costdt
=sint+C
=(x/R)/√[1+(x/R)²]+C
【参考：http://zhidao.baidu.com/question/183209510.html】

Answer 2

∫(0->L/2)dx/(x^2+R^2)^(3/2)

let
x= Rtany
dx =R(secy)^2dy
x=0 , y=0
x= L/2 , y = arctan(L/(2R))

∫(0->L/2)dx/(x^2+R^2)^(3/2)
=(1/R^2)∫(0->arctan(L/(2R))) cosy dy
= (1/R^2)[siny](0->arctan(L/(2R)))
= L/(R^2√(4R^2+L^2))

Answer 3

令x=Rtant

则原式=∫(0→arctan(L/(2R)))Rsec^2(t)dt*1/(R^3/cos^3(t))
=1/R^2∫(0→arctan(L/(2R)))costdt
=sint/R^2|(0→arctan(L/(2R)))
=L/(R^2√(L^2+4R^2))