<s>[INST] 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 $\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$?
 [/INST] 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
$$
-8\int\cos^22t\,dt=-4\int(1+\cos t)\,dt
$$
</s>