I have this indefinite integral , with $a\in \Bbb R, \: a\neq 0$
$$\int \frac{dx}{\sqrt{a^2+x^2}}, \tag 1$$
I solve the integral $(1)$ with $x=at$, and using this approach,
$$\int \frac{dx}{\sqrt{1+x^2}}=\int \frac{dx}{\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}=\cdots =\int \frac{1+\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+x^2}+x}dx=\ln\lvert \sqrt{1+x^2}+x \rvert+k, \quad k\in\Bbb R \tag 2$$
Is there a fast mode to solve the $(1)$ and the $(2)$?
On
If your high school students have learned about the hyperbolic functions, then there is a natural substitution based on the identity $\cosh^2(x)-\sinh^2(x)=1$.
Therefore, if we make the substitution $x=a\sinh(t)$ such that $dx=a\cosh(t)\,dt$, then we can write
$$\begin{align} \int\frac{1}{\sqrt{a^2+x^2}}\,dx&=\int \frac{1}{a\cosh(t)}\,a\cosh(t)\,dt\\\\ &=t+C\\\\ &=\text{arsinh}(x/a)+C\\\\ &=\frac1a \log\left( x+\sqrt{x^2+a^2}\right)+C' \end{align}$$
Hint: Substitute $x= a\tan\theta$ for $(1)$ and $x=\tan \theta$ for $(2)$.