To evaluate the indefinite integral

$\int 7^{\left(\frac{x}{2} + 1\right)} \, dx$

we can follow these steps:

Step 1: Rewrite the exponent

We express the integral in a more convenient form by breaking up the exponent:

$7^{\left(\frac{x}{2} + 1\right)} = 7^{\frac{x}{2}} \cdot 7^1 = 7 \cdot 7^{\frac{x}{2}}$

Thus, the integral becomes:

$\int 7 \cdot 7^{\frac{x}{2}} \, dx$

Step 2: Simplify the constant factor

The constant 7 can be factored out of the integral:

$7 \int 7^{\frac{x}{2}} \, dx$

Step 3: Use substitution

Let’s substitute $u = \frac{x}{2}$, which means that $du = \frac{1}{2} dx$, or equivalently $dx = 2 du$.

The integral becomes:

$7 \int 7^u \cdot 2 \, du$

This simplifies to:

$14 \int 7^u \, du$

Step 4: Integrate

Now we can integrate $7^u$. Recall that the integral of $a^u$ is:

$\int a^u \, du = \frac{a^u}{\ln(a)}$

For $a = 7$, this becomes:

$14 \cdot \frac{7^u}{\ln(7)} + C$

Step 5: Substitute back $u = \frac{x}{2}$

Finally, substitute $u = \frac{x}{2}$ back into the expression:

$14 \cdot \frac{7^{\frac{x}{2}}}{\ln(7)} + C$

Final Answer:

$\int 7^{\left(\frac{x}{2} + 1\right)} \, dx = \frac{14 \cdot 7^{\frac{x}{2}}}{\ln(7)} + C$

---

Would you like to go through any specific part of this solution in more detail?

Here are some related questions for further exploration:

1. How do you differentiate $7^{\frac{x}{2}}$?
2. What happens if the exponent had a different coefficient, like $3x$?
3. How would this integral change if we had $e^{\frac{x}{2} + 1}$ instead of $7^{\frac{x}{2} + 1}$?
4. How can you apply substitution to more complex integrals involving exponential functions?
5. What is the general form of the integral of $a^{kx}$ where $a$ and $k$ are constants?

Tip: Always look for a substitution that can simplify the exponential function's exponent to make the integration process easier.