Primitive of Root of x squared plus a squared/Logarithm Form

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Theorem

$\ds \int \sqrt {x^2 + a^2} \rd x = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } + C$


Proof

\(\ds \int \sqrt {x^2 + a^2} \rd x\) \(=\) \(\ds \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \arsinh \frac x a + C\) Primitive of $\sqrt {x^2 + a^2}$ in $\arsinh$ form
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \paren {\map \ln {x + \sqrt {x^2 + a^2} } - \ln a} + C\) $\arsinh \dfrac x a$ in Logarithm Form
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } - \frac {a^2 \ln a} 2 + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } + C\) subsuming $\dfrac {-a^2 \ln a} 2$ into the arbitrary constant

$\blacksquare$


Sources