<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] Well, i was try with this
$\int \frac{dx}{x +\sqrt{1-x^2}}=\int \frac{x-\sqrt{1-x^2}}{(x +\sqrt{1-x^2})(x -\sqrt{1-x^2})}= \int \frac{x-\sqrt{1-x^2}}{x^2-(1-x^2)}= \int \frac{x-\sqrt{1-x^2}}{2x^2-1}= \int \frac{x}{2x^2-1} -\int \frac{\sqrt{1-x^2}}{2x^2-1}$.
For the first integral we use $u=2x^2-1$, $du=4xdx$ and so on...
I believe that for the second integral we can use $t=\sqrt{1-x^2}$ then $x^2=1-t^2$, $dt=-\frac{\sqrt{1-t^2}}{t}dx$, so
$-\int \frac{\sqrt{1-x^2}}{2x^2-1}=-\int \frac{t}{2(1-t^2)-1}(\frac{-t}{\sqrt{1-t^2}})dt=\int \frac{t^2}{(1-2t^2)\sqrt{1-t^2}}= \frac{-1}{2}\int \frac{-2t^2}{(1-2t^2)\sqrt{1-t^2}}=\frac{-1}{2}\int \frac{1-2t^2 -1}{(1-2t^2)\sqrt{1-t^2}}$
$=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int \frac{dt}{(1-2t^2) \sqrt{1-t^2}}=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int \frac{dt}{(1-t^2-t^2) \sqrt{1-t^2}}=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{-1}{2}\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}$
and, again for $\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}$
If we put $u=\sqrt{1-t^2}$ then $u^3=(1-t^2)^{\frac{3}{2}} $, $t^2=1-u^2$ and $\frac{-udu}{\sqrt{1-u^2}}=dt$ implies
$\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}= \int \frac{-udu}{\sqrt{1-u^2}(u^3- (1-u^2)u)}=\int \frac{du}{(1-u^2)^{\frac{3}{2}}}$
</s>