Primitive of Inverse Hyperbolic Sine Function

Theorem

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


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:

\(\ds u\) \(=\) \(\ds \arsinh x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 {\sqrt {x^2 + 1} }\) Derivative of $\arsinh \dfrac x a$, setting $a := 1$


and let:

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


Then:

\(\ds \int \arsinh x \rd x\) \(=\) \(\ds x \arsinh x - \int x \paren {\frac 1 {\sqrt {x^2 + 1} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x \arsinh x - \int \frac {x \rd x} {\sqrt {x^2 + 1} } + C\) simplifying
\(\ds \) \(=\) \(\ds x \arsinh x - \sqrt {x^2 + 1} + C\) Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$, setting $a := 1$

$\blacksquare$


Also see


Sources

  • 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