# Integration of the Square Root of a^2+x^2

In this tutorial we shall derive the integration of the square root of a^2+x^2, and solve this integration with the help of the integration by parts methods.

The integral of $\sqrt {{a^2} + {x^2}}$ is of the form

$\int {\sqrt {{a^2} + {x^2}} dx} = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) + c$

OR
$\int {\sqrt {{a^2} + {x^2}} dx} = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}\ln \left[ {x + \sqrt {{a^2} + {x^2}} } \right] + c$

This integral can be written as
$I = \int {\sqrt {{a^2} + {x^2}} \cdot 1dx}$

Here the first function is $\sqrt {{a^2} + {x^2}}$ and the second function is $1$
$I = \int {\sqrt {{a^2} + {x^2}} \cdot 1dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Using the formula for integration by parts, we have
$\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }$

Using the formula above, equation (i) becomes
$\begin{gathered} I = \sqrt {{a^2} + {x^2}} \int {1dx – \int {\left[ {\frac{d}{{dx}}\sqrt {{a^2} + {x^2}} \left( {\int {1dx} } \right)} \right]} dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} – \int {\frac{{ – {a^2} + {a^2} + {x^2}}}{{\sqrt {{a^2} + {x^2}} }}} dx \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} – \int {\frac{{ – {a^2}}}{{\sqrt {{a^2} + {x^2}} }}} dx – \int {\frac{{{a^2} + {x^2}}}{{\sqrt {{a^2} + {x^2}} }}dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} + {a^2}\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx – \int {\sqrt {{a^2} + {x^2}} dx} \\ \end{gathered}$

Using the given integral $I = \int {\sqrt {{a^2} + {x^2}} } dx$, we get
$\begin{gathered} I = x\sqrt {{a^2} + {x^2}} + {a^2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) – I + c \\ \Rightarrow I + I = x\sqrt {{a^2} + {x^2}} + {a^2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) + c \\ \Rightarrow 2I = x\sqrt {{a^2} + {x^2}} + {a^2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) + c \\ \Rightarrow I = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) + c \\ \Rightarrow \int {\sqrt {{a^2} + {x^2}} dx} = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) + c \\ \end{gathered}$

But using the relation ${\sinh ^{ – 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$ and since we know that ${\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right)$, we have
$\Rightarrow \int {\sqrt {{a^2} + {x^2}} dx} = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}\ln \left( {x + \sqrt {{x^2} + {a^2}} } \right) + c$