Question

∫dx / [x²√(1+x²)]

Answer 1

原积分=∫[dx/[(x^*√(1+x^)]
令x=tant，则有t=arctanx， 而且有：√(1+x^)=√(1+tan^t)=√sec^t=sect;
x^=tan^t，dx=d(tant)=sec^tdt
于是，原积分化为：
∫sec^tdt/(tan^t*sect)
=∫sectdt/tan^t
=∫(1/cost)*dt/(sin^t/cos^t)
=∫cost*dt/sin^t
=∫d(sint)/sin^t
=∫(sint)^(-2) *d(sint)
=-(sint)^(-1)
=-1/sint
=-根号(1+x^2)/x
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∫1/[x^2*√(1+x^2)] dx
令x=tana
则：√(1+x^2)=seca
原式=∫1/(tan^2*seca)*sec^2a da
=∫cosa/sin^2a da
=∫1/sin^2a d(sina)
=-1/sina
=-√(1+x^2)/x

追问

=-1/sint
=-根号(1+x^2)/x
麻烦您解释下这步，谢谢！

追答

x=tant=sint/cost
x^2=(sint)^2/(cost)^2
x^2/(1+x^2)=(sint)^2
sint=x/根号(1+x^2)

追问

x^2/(1+x^2)=(sint)^2这一步还是没看懂，不好意思还是要请教您

追答

x^2=(sint)^2/(cost)^2
(cost)^2=(sint)^2/x^2
(cost)^2+(sint)^2=1
即(sint)^2/x^2+(sint)^2=1
(sint)^2=x^2/(x^2+1)

Answer 2

令x=tant dx=(sect)^2dt
原式=∫(sect)^2/[(tant)^2sect] dt
=∫sect/(tant)^2 dt
=∫cost/(sint)^2 dt
=∫cott*csct dt
=-csct+C
=-csc(arctanx)+C
=-根号(x^2+1)/x+C