Question

integration of. 1/undroot x square - 1

Answer 1

ln |x + √(x^2 - 1)| + C

Answer 2

The problem asks for the integration of $\frac{1}{\sqrt{x^2 - 1}}$. This is a standard integral form.

The general formula for the integral of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx$ is given by: $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln |x + \sqrt{x^2 - a^2}| + C$ In this specific problem, we have $a^2 = 1$, which means $a = 1$.

Substitute $a=1$ into the formula: $\int \frac{1}{\sqrt{x^2 - 1}} dx = \ln |x + \sqrt{x^2 - 1}| + C$ Here, $C$ is the constant of integration.

The domain of the integrand $\frac{1}{\sqrt{x^2 - 1}}$ requires $x^2 - 1 > 0$, which means $x^2 > 1$, or $|x| > 1$.