Primitive of Inverse Hyperbolic Sine of x over a

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Theorem

$\displaystyle \int \sinh^{-1} \frac x a \rd x = x \sinh^{-1} \frac x a - \sqrt {x^2 + a^2} + C$


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle x\) Primitive of Constant


Then:

\(\displaystyle \int \sinh^{-1} \frac x a \rd x\) \(=\) \(\displaystyle x \sinh^{-1} \frac x a - \int x \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle x \sinh^{-1} \frac x a - \int \frac {x \rd x} {\sqrt {x^2 + a^2} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle x \sinh^{-1} \frac x a - \sqrt {x^2 + a^2} + C\) Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$

$\blacksquare$


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