Evaluate the integral: I = integral sqrt(x² ± a²) dx
Evaluate the integral: $$ I = \int \sqrt{x^2 \pm a^2}\,dx. $$
Explanation of Method
Both cases are handled by integration by parts with $u=\sqrt{x^2\pm a^2}$, $dv=dx$, which produces a leftover integral proportional to the original one; solving the resulting equation algebraically for the integral gives the answer.
- Choose $u=\sqrt{x^2\pm a^2}$, $dv=dx$.
- Rewrite the leftover integral so it splits into the original integral plus a standard log-form integral.
- Solve algebraically for the integral.
Case 1 · $\int \sqrt{x^2+a^2}\,dx$
Let $u=\sqrt{x^2+a^2}$, $dv=dx$, so $du=\dfrac{x}{\sqrt{x^2+a^2}}dx$, $v=x$:
$$ I_1 = x\sqrt{x^2+a^2} - \int \frac{x^2}{\sqrt{x^2+a^2}}\,dx. $$
Writing $x^2 = (x^2+a^2)-a^2$:
$$ I_1 = x\sqrt{x^2+a^2} - \int \sqrt{x^2+a^2}\,dx + a^2\int \frac{dx}{\sqrt{x^2+a^2}}. $$
The middle integral is $I_1$ itself, so moving it to the left:
$$ 2I_1 = x\sqrt{x^2+a^2} + a^2\ln\left|x+\sqrt{x^2+a^2}\right| + C. $$
$$ I_1 = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right| + C. $$
Case 2 · $\int \sqrt{x^2-a^2}\,dx$
The identical steps, with $x^2=(x^2-a^2)+a^2$, give:
$$ I_2 = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln\left|x+\sqrt{x^2-a^2}\right| + C. $$
Summary Table
| Case | Result |
|---|---|
| $\int\sqrt{x^2+a^2}\,dx$ | $\tfrac{x}{2}\sqrt{x^2+a^2}+\tfrac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|$ |
| $\int\sqrt{x^2-a^2}\,dx$ | $\tfrac{x}{2}\sqrt{x^2-a^2}-\tfrac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|$ |
Additional Notes
- For $I_2$, the domain requires $|x|\geq a$, where $\sqrt{x^2-a^2}$ is real.
- This "self-referential" integration-by-parts trick — where the leftover integral reproduces a multiple of the original — is the same technique used for $\int e^{ax}\sin bx\,dx$ elsewhere in this collection, just applied to an algebraic rather than exponential/trigonometric integrand.
- Both results can be checked by differentiation: for $I_1$, $\tfrac{d}{dx}\left[\tfrac{x}{2}\sqrt{x^2+a^2}\right]=\tfrac{\sqrt{x^2+a^2}}{2}+\tfrac{x^2}{2\sqrt{x^2+a^2}}$, and $\tfrac{d}{dx}\left[\tfrac{a^2}{2}\ln(x+\sqrt{x^2+a^2})\right]=\tfrac{a^2}{2\sqrt{x^2+a^2}}$; summing and simplifying over the common denominator returns $\sqrt{x^2+a^2}$.
Result
The solution to the integral is: