Evaluating $\int\frac{1}{x\sqrt{x^2+1}}dx$ I am very confused by this. I am integrating the function;
$$\int\frac{1}{x\sqrt{x^2+1}}dx$$
And Wolfram alpha is telling me, the result is;
$$\log{\left(\frac{x}{\sqrt{x^2+1}+1}
\right)}$$
However, Wolfram Mathematica is telling me that the answer is;
$$\int\frac{1}{x\sqrt{x^2+1}}dx=-\mathrm{artanh}(\sqrt{x^2+1})$$
Are these two representation equivalent?
 A: As I said in the comment, the correct representation is the first one.
If we consider $f\colon (0,+\infty) \to \mathbb R$ defined as
$$
f(x)=\log{\left(\frac{x}{\sqrt{x^2+1}+1}
\right)}
$$
Then $f'(x) = \frac{1}{x\sqrt{x^2+1}}$ for every $x > 0$.
The second function
$$
g(x) =- \operatorname{arctanh}(\sqrt{x^2+1})
$$
has indead no real domain.
One explanation is that if we consider the function
$$
\tanh(x) = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}
$$
the image of  $\tanh$ is the interval $(-1,1)$ therefore its inverse function cannot be evaluated for  $\sqrt{x^2+1}$, because $\sqrt{x^2+1} \ge 1$ for every $x \in \mathbb R$,
A: Here is another representation:
$$\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{dx}{x^2\sqrt{1+\frac{1}{x^2}}} = -\operatorname{arsinh}\left(\frac{1}{x}\right)+C$$
A: Here is the short proof of logarithmic presentation of inverse hyperbolic functions (here $\operatorname{arctanh} x$):
$$
x=\tanh y=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\stackrel{z=e^y}=\frac{z^2-1}{z^2+1}\\
\implies z^2=\frac{1+x}{1-x}\implies 
y\equiv\boxed{\operatorname{arctanh} x=\frac12\log\frac{1+x}{1-x}.}
$$
A: The various representations discussed on this page are not only equivalent, but all obtainable from a substitution. You can get Wolfram Alpha's answer from $x=\tan t$, Wolfram Mathematica's from $u^2-x^2=1$ & @NinadMunshi's from $xy=1$.
