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

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## Theorem

$\displaystyle \int x^m \operatorname{sech}^{-1} \frac x a \ \mathrm d x = \begin{cases} \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{sech}^{-1} \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {a^2 - x^2} } \ \mathrm d x + C & : \operatorname{sech}^{-1} \dfrac x a > 0 \\ \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{sech}^{-1} \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {a^2 - x^2} } \ \mathrm d x + C & : \operatorname{sech}^{-1} \dfrac x a < 0 \\ \end{cases}$

## Proof

With a view to expressing the primitive in the form:

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

let:

 $\displaystyle u$ $=$ $\displaystyle \operatorname{sech}^{-1} \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac {-a} {x \sqrt{a^2 - x^2} }$ Derivative of $\operatorname{sech}^{-1} \dfrac x a$

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle x^m$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {x^{m + 1} } {m + 1}$ Primitive of Power

Then:

 $\displaystyle \int \frac {\operatorname{sech}^{-1} \dfrac x a \ \mathrm d x} {x^2}$ $=$ $\displaystyle \left({\operatorname{sech}^{-1} \frac x a}\right) \left({\frac {x^{m + 1} } {m + 1} }\right) - \int \left({\frac {x^{m + 1} } {m + 1} }\right) \left({\frac {-a} {x \sqrt{a^2 - x^2} } }\right) \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{sech}^{-1} \frac x a + \frac a {m + 1} \int \frac {x^m} {\sqrt {x^2 - a^2} } \ \mathrm d x + C$ simplifying

$\blacksquare$