The integral in the image is:

$\int (2x - 1) \sqrt{3x - 1} \, dx$

To solve this, we'll use the substitution method:

Step 1: Substitution

Let: $u = 3x - 1$

Then, differentiate $u$ with respect to $x$: $du = 3 \, dx \quad \text{or} \quad dx = \frac{du}{3}$

Also, express $2x - 1$ in terms of $u$. From $u = 3x - 1$, we can solve for $x$: $x = \frac{u + 1}{3}$

Now substitute into $2x - 1$: $2x - 1 = 2\left(\frac{u + 1}{3}\right) - 1 = \frac{2(u + 1)}{3} - 1 = \frac{2u + 2 - 3}{3} = \frac{2u - 1}{3}$

Step 2: Rewrite the Integral

Substitute into the original integral: $\int (2x - 1) \sqrt{3x - 1} \, dx = \int \frac{2u - 1}{3} \cdot \sqrt{u} \cdot \frac{du}{3}$

Simplifying: $\frac{1}{9} \int (2u - 1) \sqrt{u} \, du$

Step 3: Distribute and Integrate

Now distribute $\sqrt{u}$ to $2u - 1$: $\frac{1}{9} \int \left(2u^{3/2} - u^{1/2}\right) \, du$

Now integrate term by term:

• The integral of $2u^{3/2}$ is: $\int 2u^{3/2} \, du = \frac{2 \cdot u^{5/2}}{5/2} = \frac{4}{5} u^{5/2}$

• The integral of $u^{1/2}$ is: $\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

Step 4: Combine the Results

Substitute back into the integral: $\frac{1}{9} \left( \frac{4}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) + C$

Now substitute $u = 3x - 1$ back into the result: $\frac{1}{9} \left( \frac{4}{5} (3x - 1)^{5/2} - \frac{2}{3} (3x - 1)^{3/2} \right) + C$

This is the final answer.

Would you like more details on any step or have questions? Here are some related questions:

1. How do you handle integrals with radicals like $\sqrt{3x - 1}$?
2. What is the general method for substitution in integration?
3. How would you check the result of an integral by differentiation?
4. What are some common techniques for integrating powers of functions?
5. How does substitution simplify complex integrals?

Tip: Always check if substitution can simplify an integral, especially when you see a function and its derivative in the integrand.