integration of $\sqrt{x^2+1}$

The integral

$$I:=\int \sqrt{x^2+1}𝑑x$$

can be found by using the first Euler’s substitution (http://planetmath.org/EulersSubstitutionsForIntegration)

$$\sqrt{x^2+1}:=-x+t,$$

but another possibility is to use partial integration (http://planetmath.org/ASpecialCaseOfPartialIntegration) 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:

$$I=\int 1\cdot \sqrt{x^2+1}𝑑x=x\sqrt{x^2+1}-\int x\cdot \frac{1}{2\sqrt{x^2+1}}\cdot 2x𝑑x+C^{\prime }=x\sqrt{x^2+1}-\int \frac{x^2}{\sqrt{x^2+1}}𝑑x+C^{\prime }.$$

Writing the numerator as $(x^2+1)-1$ and dividing its minuend and subtrahend separately, we can write

$$I=x\sqrt{x^2+1}-\left(\int \sqrt{x^2+1}𝑑x-\int \frac{1}{\sqrt{x^2+1}}𝑑x\right)+C^{\prime }=x\sqrt{x^2+1}-I+\int \frac{dx}{\sqrt{x^2+1}}+C^{\prime }.$$

Having $I$ in two , 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.,

$$\int \sqrt{x^2+1}𝑑x=\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\mathrm{arsinh}x+C$$