The integral $$I \,:=\, \int\!\sqrt{x^2\!+\!1}\;dx$$ can be found by using the first Euler's substitution $$\sqrt{x^2\!+\!1} \;:=\; -x\!+\!t,$$ but another possibility is to use partial integration if one knows the integral $\int\!\frac{dx}{\sqrt{x^2+1}}$ The corresponding may be said of the more general $$\int\!\sqrt{x^2\!+\!c}\;dx.$$

We think that the integrand of $I$ has the other factor 1 and integrate partially: $$I \,=\, \int\!1\cdot\sqrt{x^2\!+\!1}\;dx \,=\, x\sqrt{x^2\!+\!1}-\!\int\!x\cdot\frac{1}{2\sqrt{x^2\!+\!1}}\cdot2x\,dx+C' \,=\, x\sqrt{x^2\!+\!1}-\!\int\!\frac{x^2}{\sqrt{x^2\!+\!1}}\,dx+C'.$$ Writing the numerator as $(x^2\!+\!1)\!-\!1$ and dividing its minuend and subtrahend separately, we can write $$ I \,=\, x\sqrt{x^2\!+\!1}-\!\left(\!\int\!\sqrt{x^2\!+\!1}\,dx-\!\int\!\frac{1}{\sqrt{x^2\!+\!1}}\,dx\right)+C' \,=\, x\sqrt{x^2\!+\!1}-I+\!\int\!\frac{dx}{\sqrt{x^2\!+\!1}}+C'.$$ Having $I$ in two places, we solve it from these equalities, obtaining $$I \,=\, \frac{x}{2}\sqrt{x^2\!+\!1}+\frac{1}{2}\!\int\frac{dx}{\sqrt{x^2\!+\!1}}+C,$$ i.e., $$\int\!\sqrt{x^2\!+\!1}\;dx \;=\; \frac{x}{2}\sqrt{x^2\!+\!1}+\frac{1}{2}\arsinh{x}+C$$