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To evaluate the integral:

\[
\int \frac{14 \, dx}{x \sqrt{1 + 2x}},
\]

we can use a substitution method to simplify the expression. Let's proceed step-by-step:

1. **Substitute**:
   Let \( u = 1 + 2x \), so that \( du = 2 \, dx \). Thus, \( dx = \frac{du}{2} \).

2. **Rewrite** the integral in terms of \( u \):
   Substituting \( x = \frac{u - 1}{2} \) and \( dx = \frac{du}{2} \), we get:
   \[
   \int \frac{14 \cdot \frac{du}{2}}{\frac{u - 1}{2} \cdot \sqrt{u}} = \int \frac{14 \cdot du}{(u - 1) \cdot \sqrt{u}}.
   \]

3. **Simplify**:
   \[
   \int \frac{14 \, du}{(u - 1) \sqrt{u}}.
   \]

4. **Separate**:
   Break the fraction into two parts to make it easier to integrate:
   \[
   \int \frac{14 \, du}{\sqrt{u}} - \int \frac{14 \, du}{u \sqrt{u}}.
   \]
   
   Simplifying each integral separately:
   \[
   \int \frac{14 \, du}{\sqrt{u}} = 14 \int u^{-\frac{1}{2}} \, du = 14 \cdot 2\sqrt{u} = 28\sqrt{u}.
   \]

   \[
   \int \frac{14 \, du}{u \sqrt{u}} = 14 \int u^{-\frac{3}{2}} \, du = 14 \cdot \left(-2u^{-\frac{1}{2}}\right) = -28u^{-\frac{1}{2}}.
   \]

5. **Combine** and back-substitute \( u = 1 + 2x \):
   \[
   28\sqrt{1 + 2x} - \frac{28}{\sqrt{1 + 2x}} + C,
   \]
   where \( C \) is the constant of integration.

Therefore, the evaluated integral is:

\[
\boxed{28\sqrt{1 + 2x} - \frac{28}{\sqrt{1 + 2x}} + C}.
\]

Let me know if you want more details or have any questions. Here are some related questions to deepen your understanding:

1. What is the importance of choosing the right substitution in integrals?
2. How do we handle integrals with square roots involving linear expressions?
3. Can you evaluate the integral using trigonometric substitution?
4. What is the geometric interpretation of this integral?
5. How does the integral change if the numerator were different?

**Tip**: Always verify your solution by differentiating it to see if you obtain the original integrand.

Evaluate the integral: \( \int \frac{14 \, dx}{x \sqrt{1 + 2x}} \).

Learn how to evaluate the integral \( \int \frac{14 \, dx}{x \sqrt{1 + 2x}} \) step-by-step using the substitution method. This problem covers essential calculus techniques and provides a clear solution, ideal for students in advanced high school or undergraduate calculus.

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