Primitive of x by Inverse Hyperbolic Cosecant of x over a

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Theorem

$\ds \int x \arcsch \frac x a \rd x = \begin {cases} \dfrac {x^2} 2 \arcsch \dfrac x a + \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x > 0 \\ \dfrac {x^2} 2 \arcsch \dfrac x a - \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x < 0 \\ \end {cases}$


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:

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


and let:

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


Then:

\(\ds \int x \arcsch \frac x a \rd x\) \(=\) \(\ds \frac {x^2} 2 \arcsch \frac x a - \int \frac {x^2} 2 \paren {\dfrac {-a} {\size x \sqrt {a^2 + x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arcsch \frac x a + \frac a 2 \int \frac {x^2 \rd x} {\size x \sqrt {a^2 + x^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arcsch \frac x a \begin {cases} \mathop + \dfrac a 2 \int \dfrac {x \rd x} {\sqrt {x^2 + a^2} } + C & : x > 0 \\ \mathop - \dfrac a 2 \int \dfrac {x \rd x} {\sqrt {x^2 + a^2} } + C & : x < 0 \end {cases}\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arcsch \frac x a \begin {cases} \mathop + \dfrac a 2 \sqrt {x^2 + a^2} + C & : x > 0 \\ \mathop - \dfrac a 2 \sqrt {x^2 + a^2} + C & : x < 0 \end {cases}\) Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
\(\ds \) \(=\) \(\ds \begin {cases} \dfrac {x^2} 2 \arcsch \dfrac x a + \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x > 0 \\ \dfrac {x^2} 2 \arcsch \dfrac x a - \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x < 0 \\ \end {cases}\) simplifying

$\blacksquare$


Also see


Sources