The integral you're asking for is:

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

Step 1: Substitution

We can simplify the integral using substitution. Let's choose:

$u = 4x + 1$

Now, differentiate both sides with respect to $x$:

$\frac{du}{dx} = 4 \quad \Rightarrow \quad dx = \frac{du}{4}$

Step 2: Rewrite the Integral

Now, substitute $u$ and $dx$ into the integral:

$\int \sqrt{4x + 1} \, dx = \int \sqrt{u} \cdot \frac{du}{4}$

We can factor out the constant $\frac{1}{4}$:

$\frac{1}{4} \int \sqrt{u} \, du$

Step 3: Integrate

Recall that the integral of $\sqrt{u} = u^{1/2}$ is:

$\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

So, the integral becomes:

$\frac{1}{4} \cdot \frac{2}{3} u^{3/2} = \frac{1}{6} u^{3/2}$

Step 4: Substitute Back $u$

Now, substitute back $u = 4x + 1$:

$\frac{1}{6} (4x + 1)^{3/2}$

Final Answer

Thus, the solution to the integral is:

$\int \sqrt{4x + 1} \, dx = \frac{1}{6} (4x + 1)^{3/2} + C$

where $C$ is the constant of integration.

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Would you like to see further details or have any questions? Here are some related topics you might find interesting:

1. How to use substitution in integrals with more complex expressions.
2. The general method for integrating square roots of polynomials.
3. Exploring integration techniques for rational expressions.
4. Understanding the role of constants in indefinite integrals.
5. The power rule and its application in integration.

Tip: Always look for a substitution that simplifies the expression inside the square root, as this often leads to a much simpler integral!