# Evaluate the integral  int (xlnx)/sqrt(x^2-1) ?

Sep 24, 2017

$\int \frac{x \ln x}{\sqrt{{x}^{2} - 1}} = \ln x \sqrt{{x}^{2} - 1} - \sqrt{{x}^{2} - 1} + \arctan \sqrt{{x}^{2} - 1} + C$

#### Explanation:

We seek:

$I = \int \frac{x \ln x}{\sqrt{{x}^{2} - 1}}$

Let $\left\{\begin{matrix}u & = \ln x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \frac{1}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = \frac{x}{\sqrt{{x}^{2} - 1}} & \implies v & = \sqrt{{x}^{2} - 1}\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

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

$\therefore I = \ln x \sqrt{{x}^{2} - 1} - \int \setminus \frac{\sqrt{{x}^{2} - 1}}{x} \setminus \mathrm{dx}$ ..... [A]

For the next integral, $\int \setminus \frac{\sqrt{{x}^{2} - 1}}{x} \setminus \mathrm{dx}$, we can perform a substitution.

Let $u = \sqrt{{x}^{2} - 1} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{x}{\sqrt{{x}^{2} - 1}}$

Then substituting intio the last integral, we get:

$\int \setminus \frac{\sqrt{{x}^{2} - 1}}{x} \setminus \mathrm{dx} = \int \setminus {u}^{2} / \left({u}^{2} + 1\right) \setminus \mathrm{du}$
$\text{ } = \int \setminus \frac{{u}^{2} + 1 - 1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\text{ } = \int \setminus 1 - \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\text{ } = u - \arctan u$

Restoring the substitution and using the result with [A] we get:

$I = \ln x \sqrt{{x}^{2} - 1} - \left\{\sqrt{{x}^{2} - 1} - \arctan \sqrt{{x}^{2} - 1}\right\} + C$
$\setminus \setminus = \ln x \sqrt{{x}^{2} - 1} - \sqrt{{x}^{2} - 1} + \arctan \sqrt{{x}^{2} - 1} + C$