The integral of $e^{\sqrt{2x+1}}$ with respect to $x$ is solved as follows:

Step 1: Substitution

Let: $u = \sqrt{2x+1}.$ Then: $u^2 = 2x + 1 \quad \text{and} \quad 2u \, du = 2 \, dx \implies dx = u \, du.$

Step 2: Rewrite the Integral

The integral becomes: $\int e^{\sqrt{2x+1}} \, dx = \int e^u u \, du.$

Step 3: Integration by Parts

We use the formula for integration by parts: $\int u e^u \, du = u e^u - \int e^u \, du.$

Here, let:

• $v = u$, so $dv = du$,
• $w = e^u$, so $dw = e^u du$.

The result of integrating by parts is: $u e^u - e^u + C,$ where $C$ is the constant of integration.

Step 4: Back-substitute $u$

Replace $u$ with $\sqrt{2x+1}$: $\int e^{\sqrt{2x+1}} \, dx = \left( \sqrt{2x+1} \cdot e^{\sqrt{2x+1}} - e^{\sqrt{2x+1}} \right) + C.$

Final Answer:

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

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Do you have any questions or would you like a step clarified?
Here are 5 related questions to explore further:

1. How is integration by parts derived and used in other integrals?
2. Can you explain how substitution simplifies certain integrals?
3. What is the meaning of $e^x$ and why does it appear frequently in calculus?
4. Can we evaluate this integral for specific bounds (definite integral)?
5. How does this integral compare to others involving square roots and exponentials?

Tip: Always verify substitution steps carefully; a small error can propagate through the solution!