Primitive of Inverse Hyperbolic Cosine Function

Theorem

$\ds \int \arcosh x \rd x = x \arcosh x - \sqrt {x^2 - 1} + C$

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:

$\ds u$

$=$

$\ds \arcosh x$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac 1 {\sqrt {x^2 - 1} }$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \arcosh x \rd x$

$=$

$\ds x \arcosh x - \int x \paren {\frac 1 {\sqrt {x^2 - 1} } } \rd x + C$

$\ds $

$=$

$\ds x \arcosh x - \int \frac {x \rd x} {\sqrt {x^2 - 1} } + C$

simplifying

$\ds $

$=$

$\ds x \arcosh x - \sqrt {x^2 - 1} + C$

$\blacksquare$

1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals