Using complex number in integration I know,
$$I_1 = \int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln|x + \sqrt{x^2-1}| + c$$
But if I factor out $i$ from the denominator, I get:
$$I_2 = -i \int \dfrac{dx}{\sqrt{1 - x^2}} = -i \sin^{-1}x + ic$$
Are these 2 expressions equivalent?
 A: Here you're starting to get into complex territory. Specifically, the $\arcsin(x)$ function for reals is defined on $|x|<1$, but your $\ln$ expression has a $\sqrt{x^2-1}$ which is defined on $|x|>1$, exactly where the normal arcsine isn't defined for real values.
Is this the end of the story? No, but you need to tread with care.

Let's say you want to make sense of $\ln(z)$ where $z$ is complex, such that $\ln(x + \sqrt{x^2 -1})$ makes sense on $|x|<1$ same as the real $\arcsin(x)$ function. Well, whenever you have a logarithmic expression like $\ln(\text{Something})$, what that expression means is that it's some value such that if you apply it to the exponential function you get the something back. In other words:
$$e^{\ln(\text{Something})} = \text{Something}$$
So what happens if we pick $\text{Something}$ to be a complex number (which is what we're looking for)? If $\text{Something} =i$ then $\ln(i)$ should be a value such that
$$
e^{\color{blue}{\ln(i)}} = i
$$
And here, by remembering Euler's formula $e^{\theta i} = \cos(\theta) + i \sin(\theta)$ we see that $e^{\color{blue}{\frac{\pi}{2}i}} = \cos\left(\frac{\pi}{2}\right)+ i \sin\left(\frac{\pi}{2}\right) = i$. So one reasonable conclusion could be that
$$
\color{blue}{\ln(i)} = \color{blue}{ \frac{\pi}{2}i}
$$
But there's a problem we glossed over. Since $\sin$ and $\cos$ are periodic, $i \frac{\pi}{2}$ is not the only value that'll work. Indeed, we see that
$$
i = e^{\color{blue}{\frac{-3\pi}{2}i}} = e^{\color{blue}{i\frac{5\pi}{2}}} = e^{\color{blue}{\frac{-7\pi}{2}i}}=e^{\color{blue}{\frac{9\pi}{2}i}} = e^{\color{blue}{\frac{-11\pi}{2}i}} = \dots
$$
And now we have an infinite possibility for values of $\ln(i)$. Here's where the idea of branch cuts comes into play, where what you do by "picking a branch" is essentially selecting one value for $\ln(i)$ out of the infinite list we showed above and we work with that as its value.

So now, let's say you pick the branch cut such that
$$
\ln(i) = \frac{\pi}{2}i
$$
This choice is usually called the "principal branch". With this choice fixed, we can even make more substantial conclusions. Notice that
$$
\ln\left(x + \sqrt{x^2-1}\right) +C \Bigg\vert_{x=0} = C+\ln(i) = C+\frac{\pi}{2}i
$$
And also
$$
-i \arcsin(x)\Bigg\vert_{x=0} =-i(0) =0
$$
So if we want these functions to be equal we must have $C+\frac{\pi}{2}i
 = 0 \implies C = -\frac{\pi}{2}i$, and thus, you can assert that
$$
\boxed{-i\arcsin(x) = \ln\left(x + \sqrt{x^2-1}\right)-\frac{\pi}{2}i, \qquad |x|<1}
$$
with $\ln(x)$ being defined on the principal branch. It's important to note that if we chose a different branch, then the constant added would need to be different, so the above equality only holds for the particular branch choice we made, but not for other choices.

tl;dr there is a way in which we can say that these functions can be equal, but one needs to be careful since when dealing with complex inputs these functions become multi-valued, so equalities need to be handled with care for them to hold any meaning.
A: Let $\sin^{-1}x = θ$
$$ \sin θ = x\\
\frac{e^{iθ} - e^{-iθ}}{2i} = x \\ $$
Solving the quadratic,
$$ e^{iθ} = ix \mp \sqrt{1 - x^2} = i(x \mp \sqrt{x^2 - 1})\\ $$
Taking log on both sides and then multiplying the numerator and denominator by the congugate,
$$\begin{align}
iθ &= \ln i + \ln |x \mp \sqrt{x^2 - 1}| \\
i \sin^{-1}x &= \ln e^{\frac{iπ}2} + \ln |\frac 1{x \pm \sqrt{x^2 - 1}}|\\
&= \frac{iπ}2 - \ln |x \pm \sqrt{x^2 - 1}|\\
\end{align}$$
So $ - i \sin^{-1}x$ is equivalent to $\ln |x ± \sqrt{x^2 - 1}|$.
