To solve the integral $\int x \sqrt{4x - 1} \, dx$, we can use the substitution method. Let's break it down step by step:

Step 1: Substitution

Let: $u = 4x - 1$ Then, the derivative of $u$ with respect to $x$ is: $du = 4dx \quad \text{or} \quad dx = \frac{du}{4}$

Step 2: Rewriting the Integral

Now we can express the integral in terms of $u$: $x = \frac{u + 1}{4}$ Substitute these into the integral: $\int x \sqrt{4x - 1} \, dx = \int \frac{u + 1}{4} \cdot \sqrt{u} \cdot \frac{du}{4}$

Simplifying: $= \frac{1}{16} \int (u + 1) \sqrt{u} \, du$ We can now expand the integrand: $= \frac{1}{16} \int (u^{3/2} + u^{1/2}) \, du$

Step 3: Integration

Now, integrate each term: $\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$ $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

Thus, the integral becomes: $\frac{1}{16} \left( \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right)$

Step 4: Substitute Back $u = 4x - 1$

Substitute $u = 4x - 1$ back into the equation: $= \frac{1}{16} \left( \frac{2}{5} (4x - 1)^{5/2} + \frac{2}{3} (4x - 1)^{3/2} \right) + C$

Simplify the constant factor: $= \frac{1}{40} (4x - 1)^{5/2} + \frac{1}{24} (4x - 1)^{3/2} + C$

Final Answer:

$\int x \sqrt{4x - 1} \, dx = \frac{1}{40} (4x - 1)^{5/2} + \frac{1}{24} (4x - 1)^{3/2} + C$

Let me know if you'd like further clarification or more steps! Here are some related questions you might find interesting:

1. How do I handle integrals with higher powers of polynomials inside square roots?
2. What are other methods of substitution for this kind of integral?
3. How do you approach integrals with mixed powers of $x$ and square roots?
4. What happens if the integrand was $\sqrt{4x - 1}$ instead of $x\sqrt{4x - 1}$?
5. How would you solve $\int \frac{1}{\sqrt{4x - 1}} \, dx$?

Tip: When integrating functions with square roots, always check if a substitution can simplify the expression into a more straightforward power of $u$.