How do you integrate ln(x+sqrt(x^2+1))?

Jun 5, 2016

By parts.

Explanation:

You can integrate it by parts with the rule

$\setminus \int f ' \left(x\right) g \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \setminus \int f \left(x\right) g ' \left(x\right)$

where we assume that

$f ' \left(x\right) = 1$ and $g \left(x\right) = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$

consequently

$f \left(x\right) = x$ and $g ' \left(x\right) = \frac{1}{\sqrt{{x}^{2} + 1}}$.

The integral is then

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

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

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