The integral shown in the image is:

$\int \frac{1}{\sqrt{4x}} \, dx$

Let’s solve it step-by-step:

Step 1: Simplify the square root

We rewrite $\sqrt{4x}$ as: $\sqrt{4x} = \sqrt{4} \cdot \sqrt{x} = 2\sqrt{x}.$ Thus, the integral becomes: $\int \frac{1}{\sqrt{4x}} \, dx = \int \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{x}} \, dx.$

Step 2: Recall the formula for $\int \frac{1}{\sqrt{x}} \, dx$

The integral of $\frac{1}{\sqrt{x}}$ is: $\int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x}.$

Step 3: Combine results

Substitute back: $\frac{1}{2} \int \frac{1}{\sqrt{x}} \, dx = \frac{1}{2} \cdot 2\sqrt{x} = \sqrt{x}.$

Step 4: Add the constant of integration

Finally, include the constant of integration $C$: $\int \frac{1}{\sqrt{4x}} \, dx = \sqrt{x} + C.$

Final Answer:

$\boxed{\sqrt{x} + C}$

Would you like a detailed explanation of any part, or have further questions? Let me know!

Related Questions:

1. How do you evaluate definite integrals involving square roots, such as $\int_{1}^{4} \frac{1}{\sqrt{4x}} \, dx$?
2. What is the general formula for $\int \frac{1}{\sqrt{ax}} \, dx$?
3. How is the substitution method used to solve similar integrals?
4. Can this integral be solved in terms of trigonometric substitution? Why or why not?
5. How can you differentiate the result $\sqrt{x} + C$ to verify the solution?

Tip:

For integrals involving square roots, try rewriting the square root in terms of powers of $x$, e.g., $\sqrt{x} = x^{1/2}$, as it often simplifies the calculation.