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

Refer to explanation

#### Explanation:

Using integration by parts we have that

int ln(x^2-1)*x'dx=x*ln(x^2-1)-intx*(2x)/(x^2-1)dx= x*ln(x^2-1)-2intx^2/(x^2-1)dx

Hence we have the integral $\int {x}^{2} / \left({x}^{2} - 1\right) \mathrm{dx}$ which can be written

as follows

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

Finally we get

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