Find the integral $\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx$ The integral can be represented as 
$$
\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx=
\int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx
$$
Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx=2\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$$
What substitution to use for solving the integral $\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$?
 A: No substitutions:
$$
\int\left(\frac{1}{\sqrt{1-x^2}}-\frac{-x}{\sqrt{1-x^2}}\right)\,dx
=\arcsin x-\sqrt{1-x^2}+c
$$
You can also do that way; continue with $u=\sqrt{t}$, so $t=u^2$ and $dt=2u\,du$; so you get
$$
\int\frac{4u^2}{(u^2+1)^2}\,du=
\int 2u\cdot\frac{2u}{(u^2+1)^2}\,du
$$
Noticing that $2u$ is the derivative of $u^2+1$ you can use integration by parts
$$
=2u\cdot\left(-\frac{1}{u^2+1}\right)-
\int2\left(-\frac{1}{u^2+1}\right)\,du
=2\arctan u-\frac{2u}{u^2+1}
$$
Do the back substitutions.
Alternative method: set $x=\cos4t$, so you have
$$
\sqrt{\frac{1+\cos4t}{1-\cos4t}}=\frac{\cos2t}{\sin2t}
$$
and the integral becomes
$$
-8\int\cos^22t\,dt=-4\int(1+\cos t)\,dt
$$
A: We can also use the substitution $u=\sqrt{1-x}$, then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}$ and $\sqrt{1+x}=\sqrt{2-u^2}$. We will also use $u=\sqrt2\sin(\theta)$
$$
\begin{align}
\int\frac{\sqrt{1+x}}{\sqrt{1-x}}\,\mathrm{d}x
&=-2\int\sqrt{2-u^2}\,\mathrm{d}u\\
&=-2\int\sqrt2\cos(\theta)\cdot\sqrt2\cos(\theta)\,\mathrm{d}\theta\\
&=-2\int(1+\cos(2\theta))\,\mathrm{d}\theta\\[3pt]
&=-2\theta-\sin(2\theta)+C\\[3pt]
&=-2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)-\sqrt{1-x^2}+C\\
&=-\arccos(x)-\sqrt{1-x^2}+C\\[6pt]
&=\arcsin(x)-\sqrt{1-x^2}+\left(C-\tfrac\pi2\right)
\end{align}
$$

Explanation of $\boldsymbol{2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)=\arccos(x)}$
Let $\alpha=\arcsin\left(\sqrt{\frac{1-x}2}\right)$, then $x=1-2\sin^2(\alpha)=\cos(2\alpha)$. Thus, $\alpha=\frac12\arccos(x)$. Therefore,
$$
2\arcsin\left(\sqrt{\frac{1-x}2}\right)=\arccos(x)
$$
