Evaluate the integral: I = integral dx/(sqrt(x²+1)
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{x^2+1}-\sqrt{x^2-1}}. $$
Explanation of Method
Rationalizing by the conjugate collapses the denominator to the constant $2$, splitting the integral into $\tfrac12\int\sqrt{x^2+1}\,dx+\tfrac12\int\sqrt{x^2-1}\,dx$, both standard radical integrals.
- Rationalize by the conjugate.
- Integrate $\int\sqrt{x^2+1}\,dx$.
- Integrate $\int\sqrt{x^2-1}\,dx$.
- Combine.
Step 1 · Rationalize
Multiplying by $\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}$, the denominator becomes $(x^2+1)-(x^2-1)=2$:
$$ I = \frac12\int\sqrt{x^2+1}\,dx + \frac12\int\sqrt{x^2-1}\,dx. $$
Step 2 · First piece
$$ \int\sqrt{x^2+1}\,dx = \frac{x\sqrt{x^2+1}}2 + \frac12\ln\left|x+\sqrt{x^2+1}\right| + C_1. $$
Step 3 · Second piece
$$ \int\sqrt{x^2-1}\,dx = \frac{x\sqrt{x^2-1}}2 - \frac12\ln\left|x+\sqrt{x^2-1}\right| + C_2. $$
Step 4 · Combine
$$ I = \frac x4\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right) + \frac14\ln\left|\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}\right| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $\sqrt{x^2+1}$ | $\tfrac x2\sqrt{x^2+1}+\tfrac12\ln|x+\sqrt{x^2+1}|$ |
| $\sqrt{x^2-1}$ | $\tfrac x2\sqrt{x^2-1}-\tfrac12\ln|x+\sqrt{x^2-1}|$ |
Additional Notes
- The domain is $|x|\geq1$.
- The result was verified by differentiation at $x=2$ and $x=3$: the derivative matches the integrand $1/(\sqrt{x^2+1}-\sqrt{x^2-1})$ exactly at both points.
Result
The solution to the integral is: