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?

Solution 1:

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

$\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}= \int \frac{-udu}{\sqrt{1-u^2}(u^3- (1-u^2)u)}=\int \frac{du}{(1-u^2)^{\frac{3}{2}}}$

Solution 2:

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.