Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form/Proof 1
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Theorem
- $\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {x + \sqrt {x^2 + a^2} } + C$
Proof
\(\ds \int \frac {\d x} {\sqrt {x^2 + a^2} }\) | \(=\) | \(\ds \arsinh {\frac x a} + C\) | Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ in $\arsinh$ form | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {x + \sqrt {x^2 + a^2} } - \ln a + C\) | $\arsinh \dfrac x a$ in Logarithm Form | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {x + \sqrt {x^2 + a^2} } + C\) | subsuming $-\ln a$ into arbitrary constant |
$\blacksquare$