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

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Theorem

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


Corollary

$\ds \int \frac {\d x} {-\sqrt {x^2 + a^2} } = \ln \size {x - \sqrt {x^2 + a^2} } + C$


Proof 1

\(\ds \int \frac {\d x} {\sqrt {x^2 + a^2} }\) \(=\) \(\ds \arsinh {\frac x a} + C\) 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

$\blacksquare$


Proof 2

Let $y^2 = a^2 + x^2$.

Then:

\(\ds 2 y \frac {\d y} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives, Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds y \frac {\d y} {\d x}\) \(=\) \(\ds x\) simplification
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} x\) \(=\) \(\ds \frac {\d x} y\)
\(\ds \) \(=\) \(\ds \frac {\d x + \d y} {x + y}\)
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\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {x^2 + a^2} }\) \(=\) \(\ds \int \frac {\d x} y\) substituting for $y$
\(\ds \) \(=\) \(\ds \frac {\d x + \d y} {x + y}\) from above
\(\ds \) \(=\) \(\ds \ln \size {x + y} + C\) Primitive of Function under its Derivative
\(\ds \) \(=\) \(\ds \ln \size {x + \sqrt {x^2 + a^2} } + C\) substituting back
\(\ds \) \(=\) \(\ds \map \ln {x + \sqrt {x^2 + a^2} } + C\) argument of $\ln$ always positive

$\blacksquare$


Also presented as

Some sources present this in the form:

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

which is the same as above, except that the constant $a$ has not been subsumed into the arbitrary constant $C$.


Also see


Sources

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