# How do you integrate  ln(x^2 -1)dx?

Jul 31, 2016

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

#### Explanation:

we can try going with IBP: $\int u v ' = u v - \int u ' v$

so

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

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

and by IBP

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

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

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

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

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

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

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

so now using partial fractions

$\frac{1}{\left(x - 1\right) \left(x + 1\right)} = \frac{1}{2} \left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)$

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

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

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