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

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

Step 1 · Trigonometric substitution

Let:

$$ x = \sec\theta \quad \text{so that} \quad dx = \sec\theta\tan\theta\,d\theta. $$

Since $\sec^2\theta-1=\tan^2\theta$:

$$ \sqrt{x^2-1} = \sqrt{\sec^2\theta-1} = \tan\theta. $$

Step 2 · Substitute into the integral

Substituting these into the integral:

$$ I = \int \frac{\sec\theta\tan\theta\,d\theta}{\sec\theta\cdot\tan\theta}. $$

Step 3 · Simplify and integrate

The $\sec\theta$ and $\tan\theta$ factors cancel completely:

$$ I = \int d\theta = \theta + C. $$

Step 4 · Substitute back

Since $x=\sec\theta$, we have $\theta=\operatorname{arcsec}x$. Thus:

$$ I = \operatorname{arcsec}x + C. $$

Summary Table

Additional Notes

• The domain requires $|x|>1$, since $\sqrt{x^2-1}$ must be real and the denominator must be nonzero.
• The secant substitution is the standard trigonometric substitution for radicals of the form $\sqrt{x^2-a^2}$; here $a=1$, so no leading constant appears in the result.
• The result can be checked by differentiating $\operatorname{arcsec}x$ and confirming it returns $\dfrac{1}{x\sqrt{x^2-1}}$ (for $x>1$).