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?
 A: 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
$\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}= \int \frac{-udu}{\sqrt{1-u^2}(u^3- (1-u^2)u)}=\int \frac{du}{(1-u^2)^{\frac{3}{2}}}$
A: To calculate this integral without trigonometric substitution, you can use a rational parameterization of the unit circle:
$$\sqrt{1-x^2}=y=\frac{2t}{1+t^2}$$
$$x=\frac{1-t^2}{1+t^2} \to dx=\frac{-4t}{(1+t^2)^2}dt$$
so your integral
$$\int \frac{d x}{x+\sqrt{1-x^{2}}}$$
becomes
$$\int \frac{4t}{t^4-2t^3-2t-1}dt$$
wich can be solved with Partial fraction decomposition.
A: 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}
A: Let $t=\frac{x}{\sqrt{1-x^2}} $, then $x=\frac{t}{\sqrt{1+t^2}}$ and $d x=\frac{1}{\left(1+t^2\right)^{\frac{3}{2}}} d t$, which transforms the integrand into a rational function.
$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} &=\int \frac{d t}{(1+t)\left(1+t^2\right)} \\
&=\frac{1}{2} \int\left(\frac{1}{1+t}+\frac{1-t}{1+t^2}\right) d t \\
&=\frac{1}{2}\left(\ln |1+t|+\int \frac{d t}{1+t^2}-\int \frac{t d t}{1+t^2}\right) \\
& =\left.\frac{1}{2}[\ln |1+t|+\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^2\right)\right]+C \\
&=\frac{1}{2}\left[\ln \left|1+\frac{x}{\sqrt{1-x^2}}\right|+ \sin ^{-1} x-\frac{1}{2} \ln \left|1+\frac{x^2}{1-x^2}\right|+C\right.\\
&= \frac{1}{2}\left[\ln \left|x+\sqrt{1-x^2}\right|+\sin ^{-1} x\right]+C
\end{aligned}
$$
A: 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.
$$
\begin{aligned}
I &=\int \frac{d x}{x+y} \\
&\left.=\frac{1}{2} \int \frac{(d x+d y)+(d x-d y)}{x+y}\right]\\&= \frac{1}{2}\left[\int \frac{d(x+y)}{x+y} + \int \frac{d(x-y)}{x+y}\right] \\ &=\frac{1}{2}\left[\ln |x+y|+\int \frac{d x}{\sqrt{1-x^2}} \right] \\ &=\frac{1}{2}\left[\ln \left| x+\sqrt{1-x^2}\right|+\sin ^{-1} x\right]+C
\end{aligned}
$$
A: $$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^2}} &\stackrel{t=\sin x}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\frac{1}{2} \int \frac{(\sin t+\cos t)+(\cos t-\sin t)}{\sin t+\cos t} d t \\
&=\frac{1}{2}\left[\int 1 d t+\int \frac{d(\sin t+\cos t)}{\sin t+\cos t}\right] \\
&=\frac{1}{2}[t+\ln |\sin t+\cos t|]+C \\
&=\frac{1}{2}\left[\sin ^{-1} x+\ln \left|x+\sqrt{1-x^2}\right|\right]+C
\end{aligned}
$$
