I plugged in the x = tan u and dx = sec^2 u du, and I got
∫[tan^2(u) + 1]^(-5/2) * sec^2(u) du
==> ∫[sec^2(u)]^(-5/2) * sec^2(u) du
==> ∫[sec^2(u)]^(-3/2) du
after integration I end up with (-3/2)(tanu/secu)
If I continue on and cancel out the cosines, I end up with (-3/2)sinu.
I don't know how to get this back in terms of x.
Wait, I think I got this.
I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped.
(-3/2)[tan(u) / √sec^2(u)]
==>(-3/2) [tanu / √(tan^2u + 1)]
==>-3x / 2√(x^2 + 1)
Also, don't forget your constant of integration. I didn't check your answer, but you can do that. If your antiderivative is correct, you should be able to differentiate it and get back the original integrand. IOW, d/dx(-3x/2sqrt(x^2 + 1)) should equal (x^2 + 1)^(-5/2).