Evaluate the integral: I = integral (sqrt(x+1)
Evaluate the integral: $$ I = \int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\,dx. $$
Explanation of Method
Multiplying by the conjugate rationalizes the denominator to the constant $2$, and expanding the resulting squared numerator collapses the integrand to the simple algebraic expression $x-\sqrt{x^2-1}$.
- Rationalize by the conjugate.
- Simplify to $x-\sqrt{x^2-1}$.
- Integrate each piece.
Step 1 · Rationalize
Multiplying by $\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}$, the denominator becomes $(x+1)-(x-1)=2$, and the numerator becomes $(\sqrt{x+1}-\sqrt{x-1})^2=2x-2\sqrt{x^2-1}$:
$$ \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} = \frac{2x-2\sqrt{x^2-1}}2 = x-\sqrt{x^2-1}. $$
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 simplified integrand from Step 1 exactly.
Result
The solution to the integral is: