請微積分高手進～～我想請問一個積分問題

∫ (z^2+x^2)^(－3/2) dx 其中z為常數
=(x/z^2)*(z^2+x^2)^(－1/2)
Sol
∫ (z^2+x^2)^(－3/2) dx
Set x=zTanw
Sinw=x/√(z^2+x^2)
dx=zSec^2 wdw
∫ (z^2+x^2)^(－3/2) dx
=∫ (z^2Sec^2 w)^(－3/2)zSec^2 wdw
=∫ [1/(z^3Sec^3 w)]zSec^2 wdw
=(1/z^2)∫ [1/Sec^3 w]Sec^2 wdw
=(1/z^2)∫ Coswdw
=(1/z^2)Sinw+c
=(1/z^2)x/√(z^2+x^2)+c

http://www.wolframalpha.com/input/?i=%E2%88%AB+%28z%5E2%2Bx%5E2%29%5E%28-3%2F2%29+dx

2011-07-13 21:13:03 補充：
Possible intermediate steps:
integral 1/(x^2+z^2)^(3/2) dx
For the integrand, 1/(x^2+z^2)^(3/2) substitute x = z tan(u) and dx = z sec^2(u) du. Then (x^2+z^2)^(3/2) = (z^2 tan^2(u)+z^2)^(3/2) = z^3 sec^3(u) and u = tan^(-1)(x/z):
= 1/z^2 integral cos(u) du

2011-07-13 21:13:12 補充：
The integral of cos(u) is sin(u):
= (sin(u))/z^2+constant
Substitute back for u = tan^(-1)(x/z):
= x/(z^2 sqrt(x^2+z^2))+constant