Primitive of Inverse Hyperbolic Sine of x over a over x squared

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Theorem

$\ds \int \frac 1 {x^2} \arsinh \dfrac x a \rd x = -\frac 1 x \arsinh \dfrac x a - \frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} }$


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \arsinh \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 {\sqrt {x^2 + a^2} }\) Derivative of $\arsinh \dfrac x a$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {x^2}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {-1} x\) Primitive of Power


Then:

\(\ds \int \frac 1 {x^2} \arsinh \dfrac x a \rd x\) \(=\) \(\ds \paren {\arsinh \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds -\frac 1 x \arsinh \dfrac x a + \int \frac {\d x} {x \sqrt {x^2 + a^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds -\frac 1 x \arsinh \dfrac x a - \frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} } + C\) Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$

$\blacksquare$


Also see


Sources

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