Integration validity of $\int\frac{1}{\sqrt{a^2 + x^2}}\,dx$ I'm just wondering if the following integration is valid.
\begin{array}{l}
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
{\rm{Let }}{u^2} = {a^2} + {x^2}\\
2udu = 2xdx\\
\frac{{du}}{x} = \frac{{dx}}{u}\\
{\rm{Let }}\frac{{du}}{x} = \frac{{dx}}{u} = A\\
du = Ax\\
dx = Au\\
\frac{{du + dx}}{{x + u}} = \frac{{Ax + Au}}{{x + u}} = A = \frac{{dx}}{u}\\
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
 = \int {\frac{1}{{\sqrt {{u^2}} }}} dx\\
 = \int {\frac{1}{u}} dx\\
 = \int {\frac{{du + dx}}{{x + u}}} \\
 = \int {\frac{{d\left( {u + x} \right)}}{{x + u}}} \\
 = \ln \left| {x + u} \right| + C\\
 = \ln \left| {x + \sqrt {{a^2} + {x^2}} } \right| + C
\end{array}
 A: Even another view on this technique could be provided via Euler substitutions. What there appeared, especially in final crucial step, was $x + u$ in the denominator, so why from the beggining assume the substitution :
$$\sqrt{a^2 + x^2} = u - x$$
Squaring both sides :
$$a^2 = u^2-2ux$$ and taking differential
$$u\mathrm{d}x=(u-x)\mathrm{d}u$$ hence
$$I = \int \frac{\mathrm{d}x}{\sqrt{a^2+x^2}} = \int \frac{(u-x)\mathrm{d}u}{(u-x)u} = \int \frac{\mathrm{d}u}{u} = \ln |u| +C = \ln\left(x+\sqrt{a^2+x^2}\right) +C $$
This approach is therefore equivallent, so what may appear "magical" is the taking multiple differentials, but this can be avoided : eliminating variable $x$ we get :
$$x = \frac{u^2-a^2}{2u}$$
Ergo $$ \sqrt{a^2+x^2} = \sqrt{a^2+\frac{(u^2-a^2)^2}{4u^2}}=\frac{u^2+a^2}{2u} $$ and for $\mathrm{d}x$ :
$$\mathrm{d}x = \mathrm{d}\left(\frac{u^2-a^2}{2u}\right)=\frac{u^2+a^2}{2u^2}\mathrm{d}u$$
Therefore
$$I = \int \frac{\mathrm{d}x}{\sqrt{a^2+x^2}} = \int \frac{\frac{u^2+a^2}{2u^2}\mathrm{d}u}{\frac{u^2+a^2}{2u}} =\int \frac{\mathrm{d}u}{u} = \ln |u| +C = \ln\left(x+\sqrt{a^2+x^2}\right) +C $$
And I was just curious what king of "magic" would arise from the third Euler substitution, clearly different than the first two, which are here somewhat equivalent, and yes here it is :
Let $$\sqrt{a^2 + x^2} = tx$$
Or $$a^2 + x^2 = t^2x^2$$
Taking differential$$2x\mathrm{d}x = 2tx^2\mathrm{d}t+2xt^2\mathrm{d}x$$
Rearanging : $$\frac{\mathrm{d}x}{tx} = \frac{\mathrm{d}t}{1-t^2}$$
Ergo
$$I=\int\frac{\mathrm{d}x}{\sqrt{a^2+x^2}}=\int\frac{\mathrm{d}x}{tx}=\int\frac{\mathrm{d}t}{1-t^2}=\frac{1}{2}\ln\left|\frac{1+t}{1-t}\right|+C=\frac{1}{2}\ln\left|\frac{x+\sqrt{a^2+x^2}}{x-\sqrt{a^2+x^2}}\right|+C$$
Indeed, after rationalisation the denominator in $\ln$ :
$$\frac{1}{2}\ln\left|\frac{x+\sqrt{a^2+x^2}}{x-\sqrt{a^2+x^2}}\right| + C = \frac{1}{2}\ln\left|\frac{\left(x+\sqrt{a^2+x^2}\right)^2}{x^2-(a^2+x^2)}\right| +C = \ln\left(x+\sqrt{a^2+x^2}\right) + C' $$
Note : Substitution $\sqrt{a^2+x^2}=a+tx$ is also possible ...
A: I played around a bit, and I was able to calculate your integral in a fashion almost identical to how you did it. I think I made it a bit more rigorous though, as I didn't manipulate the du's and dx's as if they were numbers in a fraction. Instead, I kept them "together", and didn't "break up" the derivative. 
Obviously du/dx isn't really a fraction, but the notation for derivatives and integrals often works out in a way that you can "break up" and manipulate du and dx, such that your manipulations are consistent with a more rigorous approach. Hopefully this provides some intuition as to why what you did worked:
\begin{array}{l}
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
{u^2} = {a^2} + {x^2}\\
2u\frac{{du}}{dx} = 2x\\
\frac{{1}}{u} = \frac{{1}}{x}\frac{{du}}{dx}\\
A = \frac{{1}}{u} = \frac{{1}}{x}\frac{{du}}{dx}\\
Ax = \frac{{du}}{dx}\\
Au = 1\\
\int A dx = \int \frac{1}{u} dx = \int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
\int A dx = \int \frac{{Ax + Au}}{x + u} dx=\int \frac{{\frac{{du}}{dx} + 1}}{x + u} dx \\
b = x + u\\
\frac{{db}}{dx}=1 +\frac{{du}}{dx}\\
\int \frac{{\frac{{du}}{dx} + 1}}{x + u} dx = \int \frac{\frac{db}{dx}}{b} dx \\
= ln(|b|) + C\\
= ln(|x+u|) + C\\
= ln(|x+ \sqrt {{a^2} + {x^2}}) + C\\
\end{array}
So these manipulations where you "break up" the derivative and manipulate the du's and dx's often do work. However, if you wish to keep it more rigorous and intuitive, you can instead "keep them together", and use techniques like I did, or u-substitution (for example, how this article justifies separation of variables). 
The most important point is that where you have the weird statement:
\begin{array}{l}
\int {\frac{{du + dx}}{{x + u}}} \\
\end{array}
This is as if you've multiplied the integrand by dx and 1/dx:
\begin{array}{l}
\int {\frac{{du + dx}}{{x + u}}}\frac{1}{dx}dx \\
\end{array}
And then, expanding the 1/dx into the fraction, we end up with:
\begin{array}{l}
\int {\frac{\frac{du}{dx} + 1}{{x + u}}}dx \\
\end{array}
And this of course can be evaluated using the substitution b = u + x
