Find the following integral $\int\sqrt{x^2+1} dx.$ Please complete my solution.

Question

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.

Answer 1

Hint:

$$\sqrt{x^2+1} = s-x \implies x^2+1 = s^2-2sx+x^2$$

Solve for $x$ and substitute.

Answer 2

$$s=x+\sqrt{x^2+1}\iff (s-x)^2=x^2+1\iff s^2-2sx=1\iff x=\frac{s^2-1}{2s}=\frac{s}{2}-\frac{1}{2s}. $$ So, $dx=\frac{1}{2}\left(1+\frac{1}{s^2}\right)ds$ and $$\int(x+\sqrt{x^2+1})dx=\frac{1}{2}\int s\left(1+\frac{1}{s^2}\right)ds=\frac{1}{2}\int \left(s+\frac{1}{s}\right)ds=\frac{1}{2}\left(\frac{s^2}{2}-\ln|s|\right) +C= \frac{1}{2}\left(\frac{(x+\sqrt{x^2+1})^2}{2}-\ln|x+\sqrt{x^2+1}|\right)+C. $$ Finally, $$\int\sqrt{x^2+1}dx=\int(x+\sqrt{x^2+1})dx-\frac{x^2}{2}= \frac{1}{2}\left(\frac{(x+\sqrt{x^2+1})^2}{2}-\ln|x+\sqrt{x^2+1}|-x^2\right)+C $$

Answer 3

$$\int\sqrt{x^2+1} dx=\int\frac{x^2+1}{s} ds\quad (*)$$

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 $(*)$.