How to find substitution for integral: $\int \frac{1}{\sqrt{x^2+a}} dx$? How to find substitution for integral: $$\int\frac{1}{\sqrt{x^2+a}} dx$$ (a<>0).
On Integral Calculator, they gave me this for first substitution:

Substitute $$u=\frac{x}{\sqrt{a}}\longrightarrow\ \frac{du}{dx}=\frac{1}{\sqrt{a}}\ (steps)\ ⟶\ dx=\sqrt{a}du:
=\int\frac{\sqrt{a}}{\sqrt{au2+adu}}$$

but how I could find that $$\frac{x}{\sqrt{a}}$$ substitution?

What you posted looks like a first step that won't get you very far anyway. Here is a different approach.
You are motivated to make $x^2+a$ into a perfect square. This resembles the trig identity:
$$\tan^2t+1=\sec^2t$$
Except you need to multiply by $a$ to get that $1$ to become an $a$:
$$a\tan^2t+a=a\sec^2t$$
$$\left(\overbrace{\sqrt{a}\tan t}^{x}\right)^2+a=\left(\sqrt{a}\sec t\right)^2$$
So I suggest you substitute $x=\sqrt{a}\tan t$, $dx=\sqrt{a}\sec^2 t$, and see where that takes you.
