Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} \\\\
&= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2\tan^2\theta}} = \int \sec\theta\\\\
&= \ln|\sec\theta + \tan\theta| + C = \ln| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}|
\end{align}
$$
However, wolphram alpha says the answer is $\ln |x+ \sqrt{x^2-9}$
I am wondering how did it get rid of the 3 in the denominator? This is pretty much how I got my answer:
$$
x = 3\sec\theta \\
\frac{x}{3} = \sec\theta \\
\frac{\sqrt{x^2-9}}{3} = \tan\theta
$$
 A: Note that
$$\ln\left| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| = \ln\left|x+\sqrt{x^2-9}\right|-\ln 3$$
and $\ln3$ is a constant. Since $C$ is an arbitrary constant, you can define a new constant $C_1 = C - \ln3$ and the answer is
$$\ln\left|x+\sqrt{x^2-9}\right| + C_1$$
A: Another approach:
$$I=\int\frac{dx}{\sqrt{x^2-9}}=\frac{1}{3}\int\frac{dx}{\sqrt{\left(\frac{x}{3}\right)^2-1}}$$
Let $u=\frac{x}{3}\implies3du=dx\implies$
$$I=\int\frac{du}{\sqrt{u^2-1}}=\text{arccosh}(u)+C=\text{arccosh}\left(\frac{x}{3}\right)+C$$
But recall the identity for $x\in[1,\infty)$ $$\text{arccosh}(x)=\ln\left(x+\sqrt{x^2-1}\right)$$
So $$\text{arccosh}\left(\frac{x}{3}\right)+C=\ln\left(\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}\right)$$
Let $C=-\ln(3)$ and we have the Wolfram approach. 
A: The indefinite integral of $f$, i.e. $\int f$ is not a function, but a set of functions:
$$\int f := \{g| g'=f \}$$
But we usually denote the set with one of it's element $+C$:
$$\int f = g + C \tag{1}$$
Or simply without the $+C$. Clearly, the $=$ sign in $(1)$ is not the same you'd write in $1=1$, for example. A really precise mathematician would use a different symbol, for example:
$$\int f \color{blue}{=} g$$
with $\color{blue}{=}$ means that $a\color{blue}{=}b$ iff $a'=b'$. So they would write:
$$\int x \mathrm{d}x \color{blue}{=} \frac{x^2}{2} \color{blue}{=} \frac{x^2}{2}+3$$
But we are lazy, and usually the meaning of $=$ is clear from the context.
