1. **State the problem:** We need to evaluate the integral $$\int e^x \sqrt{81 - e^{2x}} \, dx.$$\n\n2. **Substitution:** Let $$u = e^x.$$ Then, $$du = e^x dx = u dx \implies dx = \frac{du}{u}.$$\n\n3. **Rewrite the integral in terms of $u$:**\n$$\int e^x \sqrt{81 - e^{2x}} \, dx = \int u \sqrt{81 - u^2} \cdot \frac{du}{u} = \int \sqrt{81 - u^2} \, du.$$\n\n4. **Simplify the integral:** The integral reduces to $$\int \sqrt{81 - u^2} \, du,$$ which is a standard form representing the area under a semicircle of radius 9.\n\n5. **Use the formula for the integral:**\n$$\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \arcsin\left(\frac{u}{a}\right) + C,$$ where $$a = 9.$$\n\n6. **Apply the formula:**\n$$\int \sqrt{81 - u^2} \, du = \frac{u}{2} \sqrt{81 - u^2} + \frac{81}{2} \arcsin\left(\frac{u}{9}\right) + C.$$\n\n7. **Back-substitute $u = e^x$:**\n$$\int e^x \sqrt{81 - e^{2x}} \, dx = \frac{e^x}{2} \sqrt{81 - e^{2x}} + \frac{81}{2} \arcsin\left(\frac{e^x}{9}\right) + C.$$\n\n**Final answer:**\n$$\boxed{\frac{e^x}{2} \sqrt{81 - e^{2x}} + \frac{81}{2} \arcsin\left(\frac{e^x}{9}\right) + C}.$$