To solve the integral $\int e^{\sqrt{x}} \, dx$, let's proceed step by step.

Step 1: Substitution

Let $u = \sqrt{x}$. Then, $x = u^2$, and differentiate both sides to find $dx$: $dx = 2u \, du$

Step 2: Rewrite the Integral

Substituting $u = \sqrt{x}$ and $dx = 2u \, du$ into the integral: $\int e^{\sqrt{x}} \, dx = \int e^u \cdot 2u \, du$ This simplifies to: $2 \int u e^u \, du$

Step 3: Integration by Parts

Now, apply integration by parts to $\int u e^u \, du$. Let: $v = u \quad \text{and} \quad dw = e^u \, du$ Then, differentiate and integrate: $dv = du \quad \text{and} \quad w = e^u$ Using the integration by parts formula $\int v \, dw = vw - \int w \, dv$: $\int u e^u \, du = u e^u - \int e^u \, du$ $= u e^u - e^u + C$ Now, substitute back into the original expression: $2 \int u e^u \, du = 2 \left( u e^u - e^u \right) + C$

Step 4: Substitute Back $u = \sqrt{x}$

Finally, substitute $u = \sqrt{x}$ back into the equation: $= 2 \left( \sqrt{x} \cdot e^{\sqrt{x}} - e^{\sqrt{x}} \right) + C$ $= 2 e^{\sqrt{x}} \left( \sqrt{x} - 1 \right) + C$

Final Answer

$\int e^{\sqrt{x}} \, dx = 2 e^{\sqrt{x}} \left( \sqrt{x} - 1 \right) + C$

Would you like more details on any part of this solution? Here are some related questions to consider:

1. How does substitution work in integration?
2. What is the integration by parts formula?
3. How can integration by parts be used in different integrals?
4. What are the common substitutions used in integrals involving square roots?
5. How does one handle integrals involving exponential functions?

Tip: When dealing with integrals that contain compositions of functions like $e^{\sqrt{x}}$, substitution is often the first method to try!