Evaluate the integral: I = integral dx/(x- sqrt(x²-1))

Evaluate the integral: $$ I = \int \frac{dx}{x-\sqrt{x^2-1}}. $$

Explanation of Method

Multiplying by the conjugate $x+\sqrt{x^2-1}$ rationalizes the denominator to $1$, collapsing the integrand to the simple expression $x+\sqrt{x^2-1}$.

Rationalize by the conjugate.
Integrate the resulting expression.

Step 1 · Rationalize

$$ \frac1{x-\sqrt{x^2-1}}\cdot\frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} = x+\sqrt{x^2-1}. $$

Step 2 · Integrate

Using $\int\sqrt{x^2-1}\,dx=\tfrac x2\sqrt{x^2-1}-\tfrac12\ln|x+\sqrt{x^2-1}|$:

$$ I = \int x\,dx + \int\sqrt{x^2-1}\,dx = \frac{x^2}2 + \frac x2\sqrt{x^2-1} - \frac12\ln\left|x+\sqrt{x^2-1}\right| + C. $$

Summary Table

Piece	Antiderivative
$x$	$x^2/2$
$\sqrt{x^2-1}$	$\tfrac x2\sqrt{x^2-1}-\tfrac12\ln\|x+\sqrt{x^2-1}\|$

Additional Notes

The domain requires $|x|\geq1$.
The result was verified by differentiation: $\dfrac{d}{dx}\left[\dfrac{x^2}2+\dfrac x2\sqrt{x^2-1}-\dfrac12\ln|x+\sqrt{x^2-1}|\right] = x+\sqrt{x^2-1}$, matching the rationalized integrand from Step 1 exactly.

Result

The solution to the integral is:

$$ I = \frac{x^2+x\sqrt{x^2-1}}2 - \frac12\ln\left|x+\sqrt{x^2-1}\right| + C. $$