Integral with negative radicand $ \int \frac{dx}{\sqrt{-a^2-x^2}}$ I saw this integral:
$$
\int \frac{dx}{\sqrt{-a^2-x^2}}.
$$
I tried to solve it by taking $-1$ as common factor which leaded me to:
$\frac{1}{i\sqrt{x^2+a^2}}$ which leads to:
$-i\ln(\sqrt{x^2+a^2}+x)$
However I placed it on Wolfram Alpha and it says the result is:
$$
\arctan\left(\frac{x}{\sqrt{-a^2-x^2}}\right)
$$
So my question would be..is it possible that both expressions are equivalent??? Sorry if this a dumb question....
Thanks in advance.

Note $\arctan (i z) = i \text{arctanh(}z) =i \ln \sqrt{\frac{1+z}{1-z}}$.
Thus
$$
\arctan\left(\frac{x}{\sqrt{-a^2-x^2}}\right)
=\arctan\left(-i \frac{x}{\sqrt{a^2+x^2}}\right)\\
= -i \ln \sqrt{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}-x}}
= -i \ln {\frac{\sqrt{a^2+x^2}+x}{a^2}}\\
\hspace{-4cm}=-i\ln(\sqrt{x^2+a^2}+x)+C$$
