Primitive of Power of x by Inverse Hyperbolic Secant of x over a

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Theorem

$\ds \int x^m \arsech \frac x a \rd x = \dfrac {x^{m + 1} } {m + 1} \arsech \dfrac x a + \dfrac 1 {m + 1} \int \dfrac {x^m} {\sqrt {a^2 - x^2} } \rd x + C$

where $\arsech$ denotes the real area hyperbolic secant.


Corollary

$\ds \int x^m \paren {-\sech^{-1} \frac x a} \rd x = -\dfrac {x^{m + 1} } {m + 1} \paren {-\sech^{-1} \frac x a} \dfrac x a - \dfrac 1 {m + 1} \int \dfrac {x^m} {\sqrt {a^2 - x^2} } \rd x + C$

where $-\sech^{-1}$ denotes the negative branch of the real inverse hyperbolic secant multifunction.


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:

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


and let:

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


Then:

\(\ds \int \frac {\arsech \dfrac x a \rd x} {x^2}\) \(=\) \(\ds \paren {\arsech \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arsech \frac x a + \frac a {m + 1} \int \frac {x^m} {\sqrt {x^2 - a^2} } \rd x + C\) simplifying

$\blacksquare$


Also see


Sources

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