<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] Let $y=\sqrt{1-x^2}$, then $ydy=-xdx$.
Noticing that
$$d(x-y)= dx-\left(-\frac{x}{y} d x\right)=\frac{(x+y)dx}{y} \Rightarrow  \frac{d(x-y)}{x+y}=\frac{d x}{\sqrt{1-x^2}}, $$
we split the integral into 2 parts as below.
$$
\begin{aligned}
I &=\int \frac{d x}{x+y} \\
&\left.=\frac{1}{2} \int \frac{(d x+d y)+(d x-d y)}{x+y}\right]\\&= \frac{1}{2}\left[\int \frac{d(x+y)}{x+y} + \int \frac{d(x-y)}{x+y}\right] \\ &=\frac{1}{2}\left[\ln |x+y|+\int \frac{d x}{\sqrt{1-x^2}} \right] \\ &=\frac{1}{2}\left[\ln \left| x+\sqrt{1-x^2}\right|+\sin ^{-1} x\right]+C
\end{aligned}
$$
</s>