I agree with topsquark, trig substitution seems like the way to go.
you have int(16+x^2)^(-3/2) = int((4)^2+x^2)^(-3/2) = int[1/(sqrt((4)^2 + x^2)^3].
trig substitution says, if your integral contains sqrt(a^2 + x^2), where a is a constant, you use the sustitution, x = atan@ and the identity 1 + tan^2@ = sex^2@
Let x = 4*tan(t) ==> dx = 4*sec^2(t) dt
Then
Int[(16+x^2)^(-3/2)dx] =
Int[(16+16{tan^2(t)})^(-3/2) * 4sec^2(t) dt]
= 4*(16)^(-3/2)*Int[(1 + tan^2(t))^(-3/2) sec^2(t) dt]
= 4*(1/64)*Int[(sec^2(t))^(-3/2) sec^2(t) dt]
= (4/64)*Int[sec^2(t)/sec^3(t) dt]
= (1/16)*Int[1/sec(t) dt]
= (1/16)*Int[cos(t) dt]
= (1/16)*sin(t)
Now, t = tan^{-1}(x/4), thus
Int[(16+x^2)^(-3/2)dx] = (1/16)*sin[tan^{-1}(x/4)]
(plus some constant of integration.)
Now, imagine a right triangle with legs x and 4, then the hypotenuse will be sqrt(x^2 + 16). We have an angle t (which is tan^{-1}(x/4)) across from the side of length x. Then the sine of this angle will be: x/sqrt(x^2 + 16). So
sin[tan^{-1}(x/4)] = x/sqrt(x^2 + 16)
So finally:
Int[(16+x^2)^(-3/2)dx] = (1/16)*x/sqrt(x^2 + 16)
(Note: I finally found the error in my original post. This is now correct.)
-Dan