<s>[INST] 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?
 [/INST] $$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^2}} &\stackrel{t=\sin x}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\frac{1}{2} \int \frac{(\sin t+\cos t)+(\cos t-\sin t)}{\sin t+\cos t} d t \\
&=\frac{1}{2}\left[\int 1 d t+\int \frac{d(\sin t+\cos t)}{\sin t+\cos t}\right] \\
&=\frac{1}{2}[t+\ln |\sin t+\cos t|]+C \\
&=\frac{1}{2}\left[\sin ^{-1} x+\ln \left|x+\sqrt{1-x^2}\right|\right]+C
\end{aligned}
$$
</s>