How do you find the integral ln( x + sqrt(x^2 -1)) ?

Mar 14, 2018

$\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx} = x \ln \left(x + \sqrt{{x}^{2} - 1}\right) - \sqrt{{x}^{2} - 1} + C$

Explanation:

Use a hyperbolic substitution.

Put:

$x = \cosh u$

Then:

$\frac{\mathrm{dx}}{\mathrm{du}} = \sinh u$

and:

$u = \ln \left(x + \sqrt{{x}^{2} - 1}\right)$

and:

$\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx} = \int \ln \left(\cosh u + \sqrt{{\cosh}^{2} u - 1}\right) \frac{\mathrm{dx}}{\mathrm{du}} \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = \int \ln \left(\cosh u + \sqrt{{\sinh}^{2} u}\right) \sinh u \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = \int \ln \left(\cosh u + \sinh u\right) \sinh u \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = \int \ln \left({e}^{u}\right) \sinh u \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = \int u \sinh u \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = \int \left(u \sinh u + \cosh u\right) - \cosh u \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = u \cosh u - \sinh u + C$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} - 1}\right) \mathrm{dx}} = x \ln \left(x + \sqrt{{x}^{2} - 1}\right) - \sqrt{{x}^{2} - 1} + C$