# The integral of sqrt(x^2+1) dx

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Let us look at

$\int {\sqrt {{x^2} + 1} \,dx}$

Using the result from the end of the last page we get

$\begin{gathered} \int {\sqrt {{x^2} + 1} \,dx} = \int {\sqrt {{x^2} - {i^2}} \,dx} \hfill \\ \quad = \frac{1}{2}\sqrt {{x^2} - {i^2}} - \frac{1}{2}{i^2}\ln \left( {x + \sqrt {{x^2} - {i^2}} } \right) + C \hfill \\ \quad = \frac{1}{2}x\sqrt {{x^2} + 1} + \frac{1}{2}\ln \left( {x + \sqrt {{x^2} + 1} } \right) + C \hfill \\ \end{gathered}$

Oh – that was quick. Let us also have a look at the graph of the integrand vs. the integral.

Up a level : Integrals
Previous page : The integral of sqrt(x^2-1) dx
Next page : Differentiation under the integral sign - Leibniz integral ruleLast modified: May 18, 2019 @ 10:52