Find the integral of (Ar/(r^2+x^2)^(3/2))dx from a to infinity

where A, r, and a are all constants.

I set x=rtan(theta), so x^2=r^2tan(theta)^2, dx=rsec(theta)^2(dtheta), and theta=arctan(x/r)
then Ar(rsec(theta)^2(dtheta))/(r^2+r^2tan(th…
which simplifies to A*integral of 1/sec(theta)(dtheta) from a to infintiy
which pans out to be A[-sin(arctan(x/r))] from a to infinity
equals A[-1+sin(arctan(a/r))]

Which apparently is incorrect. Where did I go wrong? Thanks in advance for the help.

I'm just going to rewrite all of this because it's rather difficult to read... I'll also show you a little trick along the way.

x = r tan(t), dx = r sec²(t) dt

Change the limits right now, it will make the whole thing a lot easier later.
a = r tan(t), t = arctan(a/r) = b (just for simplicity)
∞ = r tan(t), t = arctan(∞) = π/2 (I apologize for the informality, this should be done with a limit)

Then we have:
Ar² ∫ sec²(t) dt / r³(1 + tan²(t))^(3/2)
A/r ∫ dt / sec(t)
A/r ∫ cos(t) dt
A/r sin(t)

Now just evaluate at the limits:
A/r [1 - sin(arctan(a/r))]
(A/r)[1 - (a / √(a² + r²))]


So an r got missed in the denominator, and the composition of trig with inverse trig becomes algebraic.