<s>[INST] Find the following integral $\int\sqrt{x^2+1} dx.$ Please complete my solution. I am reading a calculus book.
There is the following problem with a hint:

Find the following integral $$\int\sqrt{x^2+1} dx.$$
Hint: use the substitution $x+\sqrt{x^2+1}=s$.

I was not able to complete my solution.
My attempt is here:

$s=x+\sqrt{x^2+1}$.
$ds=(1+\frac{x}{\sqrt{x^2+1}})dx=\frac{s}{\sqrt{x^2+1}}dx$.
$dx=\frac{\sqrt{x^2+1}}{s}ds$.
So, $\int\sqrt{x^2+1} dx=\int\frac{x^2+1}{s} ds$.

Please complete my solution.
 [/INST] $$\int\sqrt{x^2+1} dx=\int\frac{x^2+1}{s} ds\quad (*)$$
Right.
$\ s=x+\sqrt{x^2+1}.\ $ If we expand $\ s^2\ $ we get,
$$s^2 = 2xs+1\implies s^2-1=2xs\implies 2x=\frac{s^2-1}{s}$$
and continue trying to find an expression for $x^2$, then substitute into $(*)$.
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