有关大学数学的不定积分，求助！

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x^2-2x = (x-1)^2 -1

let
x-1 = secu
dx = secu.tanu du
x=2, u=0
x=+∞ , u=π/2
∫(2->+∞) dx/[(x-1)^4.√(x^2-2x)]
=∫(0->π/2) secu.tanu du/[(secu)^4.tanu]
=∫(0->π/2) (cosu)^3 du
=∫(0->π/2) (cosu)^2 dsinu
=∫(0->π/2) [ 1-(sinu)^2] dsinu
=[ u -(1/3)(sinu)^2]|(0->π/2)
= π/2 - 1/3

这是反常积分，得换成一般变量，然后求极限，最后有点别扭，主要是cos为0的，sin为1，cos为1的，sin为0的转换.

x^2 - 2x = (x-1)^2 - 1, 令 x-1 = secu， 则 x = 1+secu， dx = secutanudu
I = ∫<2, +∞>dx/[(x-1)^4 √(x^2-2x)] = ∫<0, π/2>secutanudu/[(secu)^4 tanu]
= ∫<0, π/2>du/(secu)^3 = ∫<0, π/2>(cosu)^3du = ∫<0, π/2>[1-(sinu)^2]dsinu
= [sinu-(sinu)^3/3]<0, π/2> = 2/3