The integral provided is:

$\int \sqrt{e^{5x}} \, dx$

To solve this using the substitution method (let $u$-substitution), let's proceed with the following steps:

Step 1: Simplify the expression

We can rewrite the integrand $\sqrt{e^{5x}}$ as:

$\sqrt{e^{5x}} = \left(e^{5x}\right)^{1/2} = e^{\frac{5x}{2}}$

So, the integral becomes:

$\int e^{\frac{5x}{2}} \, dx$

Step 2: Apply substitution

Let:

$u = \frac{5x}{2}$

Then:

$\frac{du}{dx} = \frac{5}{2} \quad \Rightarrow \quad dx = \frac{2}{5} du$

Step 3: Rewrite the integral in terms of $u$

Substitute $u = \frac{5x}{2}$ and $dx = \frac{2}{5} du$:

$\int e^{u} \cdot \frac{2}{5} \, du$

Now, factor out the constant $\frac{2}{5}$:

$\frac{2}{5} \int e^{u} \, du$

Step 4: Integrate

The integral of $e^u$ is simply $e^u$:

$\frac{2}{5} e^u + C$

Step 5: Substitute $u$ back in terms of $x$

Recall that $u = \frac{5x}{2}$, so:

$\frac{2}{5} e^{\frac{5x}{2}} + C$

Thus, the solution to the integral is:

$\frac{2}{5} e^{\frac{5x}{2}} + C$

Would you like further details on any step or have any other questions?

Here are 5 questions you might consider based on this:

1. What happens if the exponent is more complex, such as $e^{5x^2}$?
2. How can we apply substitution in more complicated integrals involving trigonometric functions?
3. What is the difference between $u$-substitution and integration by parts?
4. Can this method be used if the function inside the square root was more complex, like $\sqrt{e^{5x} + x}$?
5. How would you evaluate a definite integral for this same problem, say from $x = 0$ to $x = 1$?

Tip: Always remember to simplify the integrand first, if possible, to make substitution easier.