Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$. 
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$

An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
 &=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without  trigonometric substitution?

Let $y= x+\sqrt{1-x^2}$. Then, $dx=\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x}dy$ and
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} 
& =\int \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x} \frac{dy}y\\
 &=\ \frac12\int \bigg( \frac1y + \frac{1}{\sqrt{1-x^2}-x}\bigg)dy\\
 &=\ \frac12\int \frac{dy}y+\frac12 \int  \frac{dx}{\sqrt{1-x^2}}\\
&= \frac12 \ln y +\frac12\sin^{-1}x+C
\end{aligned}
