I am struggling to see how two antiderivatives of the same function—obtained in two different ways—are equivalent (what I mean by equivalent is that they differ from just a constant), if they even are equivalent.

The function in question is

$$
begin{align}
f colon quad {xinmathbb{R} mid e^{2x} ge 9} &tomathbb{R}\
x &mapsto f(x) = frac{1}{sqrt{e^{2x} - 9}}
end{align},.
$$

Wanting to find $I = displaystyleint f(x)mathrm{d}x$, I proceeded as follows.

$$
int frac{1}{sqrt{e^{2x} - 9}}mathrm{d}x = frac{1}{3} int frac{1}{sqrt{left(dfrac{e^x}{3}right)^2 - 1}}mathrm{d}x,.
$$

Since $left(frac{e^x}{3}right)^2 in[1,+infty]$ and the image of $cosh$ is $[1, +infty)$,
$$exists t in mathbb{R}colonquad left(dfrac{e^x}{3}right)^2 = cosh^2t\
Rightarrow mathrm{d}x = tanh t mathrm{d}t,.$$

Thus,

$$
I = frac{1}{3} int frac{tanh t}{sqrt{cosh^2 t - 1}}mathrm{d}t = frac{1}{3}int mathrm{sech} t mathrm{d}t,.
$$

Here is where the two different ways diverge. The first thing I tried was using the hyperbolic function's definition.

$$
I = frac{1}{3}int frac{2e^t}{e^{2t} + 1} mathrm{d}t,;
qquadtext{let}quad
begin{cases}
u = e^t\
mathrm{d}u = e^tmathrm{d}t
end{cases}\ \
Rightarrow I = frac{2}{3} int frac{1}{u^2 + 1} mathrm{d}u = frac{2}{3} arctan u + C = frac{2}{3} arctan e^t + C = \
= frac{2}{3} arctan e^{text{arccosh} frac{e^x}{3}} + C = frac{2}{3} arctan e^{lnleft(frac{e^x + sqrt{e^{2x}-9}}{3}right)} + C =\
= frac{2}{3} arctan left(frac{e^{x} + sqrt{e^{2x}-9}}{3}right) + C,.
$$

The alternative solution is

$$
I = frac{1}{3}int mathrm{sech} t mathrm{d}t = frac{1}{3}int frac{cosh t}{cosh^2 t} mathrm{d}t = frac{1}{3}int frac{cosh t}{1 +sinh^2 t} mathrm{d}t =\
= frac{1}{3}arctan left(sinh tright) + C = frac{1}{3}arctan left(sinh left(mathrm{arccosh} frac{e^x}{3}right)right) + C = frac{1}{3}arctan frac{sqrt{e^{2x} - 9}}{3} + C,.
$$

I don't see any obvious way in which $$frac{2}{3} arctan left(frac{e^x + sqrt{e^{2x}-9}}{3}right) + C$$ and $$frac{1}{3}arctan frac{sqrt{e^{2x} - 9}}{3} + C$$ are equivalent antiderivatives, although they do seem to be.

Update

Following Yuriy S's suggestion, I tried the following. Let
$$
s = frac{sqrt{e^{2x} - 9}}{3},.
$$
Thus,
$$
arctan s = 2arctan frac{sqrt{1+s^2}-1}{s} = 2arctan frac{sqrt{1+frac{e^{2x}-9}{9}}-1}{frac{sqrt{e^x - 9}}{3}} = 2arctan frac{3e^x-3}{sqrt{e^x - 9}},.
$$

On the other hand, let
$$
r = frac{e^x + sqrt{e^{2x} - 9}}{3},,
$$
so that
$$
frac{1}{r} = frac{3}{e^x + sqrt{e^{2x} - 9}} cdot frac{e^x - sqrt{e^{2x} - 9}}{e^x - sqrt{e^{2x} - 9}} = frac{e^x - sqrt{e^{2x} - 9}}{3},.
$$

I was striving to apply the formula
$$
arctan x + arctan frac{1}{x} = pmfrac{pi}{2}
$$
but I got stuck here.

Whenever you face this problem, you can check the answer by differentiating. If
$$
F(x)=frac{2}{3} arctan left(frac{e^{x} + sqrt{e^{2x}-9}}{3}right) + C
$$
then
begin{align}
F'(x)
&=frac{2}{3}frac{1}{1+left(dfrac{e^{x} + sqrt{e^{2x}-9}}{3}right)^2}
frac{1}{3}left(e^x+frac{e^{2x}}{sqrt{e^{2x}-9}}right) \[4px]
&=frac{2}{9+e^{2x}+e^{2x}-9+2e^xsqrt{e^{2x}-9}}
frac{e^x(e^x+sqrt{e^{2x}-9},)}{sqrt{e^{2x}-9}} \[4px]
&=frac{1}{sqrt{e^{2x}-9}}
end{align}
We can try differentiating the second function
$$
G(x)=frac{1}{3}arctanfrac{sqrt{e^{2x}-9}}{3}+C
$$
and we get
begin{align}
G'(x)
&=frac{1}{3}frac{1}{1+left(dfrac{sqrt{e^{2x}-9}}{3}right)^2}
frac{1}{3}frac{e^{2x}}{sqrt{e^{2x}-9}} \[4px]
&=frac{1}{9+e^{2x}-9}frac{e^{2x}}{sqrt{e^{2x}-9}} \[4px]
&=frac{1}{sqrt{e^{2x}-9}}
end{align}

Good job in both cases.

You can easily compute the constant difference by the limits at $infty$ (with $C=0$ in both cases):
$$
lim_{xtoinfty}F(x)=frac{2}{3}frac{pi}{2}=frac{pi}{3}
qquad
lim_{xtoinfty}G(x)=frac{1}{3}frac{pi}{2}=frac{pi}{6}
$$

Alternative solution: substitute $sqrt{e^{2x}-9}=3t$, so
$$
x=frac{1}{2}log(9(t^2+1))
$$
and
$$
dx=frac{t}{t^2+1},dt
$$
so the integral becomes
$$
intfrac{1}{t^2+1},dt=arctan t+C=arctanfrac{sqrt{e^{2x}-9}}{3}+C
$$