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