Let's proceed to evaluate the integral you provided:

$\int \frac{x+1}{\sqrt{3x-2}} \, dx$

Using the substitution $u = 3x - 2$, we will follow the steps below.

Step 1: Substitution

Let: $u = 3x - 2$

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

Step 2: Express everything in terms of $u$

Now, express $x+1$ in terms of $u$. From the substitution $u = 3x - 2$, we can solve for $x$: $x = \frac{u + 2}{3}$ Thus: $x + 1 = \frac{u + 2}{3} + 1 = \frac{u + 5}{3}$

Also, $\sqrt{3x - 2} = \sqrt{u}$.

Step 3: Rewrite the integral

Substitute these expressions into the original integral:

$\int \frac{x+1}{\sqrt{3x-2}} \, dx = \int \frac{\frac{u + 5}{3}}{\sqrt{u}} \cdot \frac{du}{3}$

This simplifies to:

$\frac{1}{9} \int \frac{u + 5}{\sqrt{u}} \, du$

Step 4: Split the integral

Now, split the integrand into two parts:

$\frac{1}{9} \int \frac{u}{\sqrt{u}} \, du + \frac{1}{9} \int \frac{5}{\sqrt{u}} \, du$

This simplifies further:

$\frac{1}{9} \int u^{1/2} \, du + \frac{5}{9} \int u^{-1/2} \, du$

Step 5: Integrate each term

Now, integrate each term:

1. For the first integral: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

2. For the second integral: $\int u^{-1/2} \, du = 2u^{1/2}$

Step 6: Combine the results

Now, substitute the results back into the integral:

$\frac{1}{9} \left( \frac{2}{3} u^{3/2} + \frac{5}{2} u^{1/2} \right)$

This simplifies to:

$\frac{2}{27} u^{3/2} + \frac{5}{18} u^{1/2}$

Step 7: Substitute $u = 3x - 2$ back

Now, substitute $u = 3x - 2$ back into the result:

$\frac{2}{27} (3x - 2)^{3/2} + \frac{5}{18} (3x - 2)^{1/2} + C$

So the final answer is:

$\int \frac{x+1}{\sqrt{3x-2}} \, dx = \frac{2}{27} (3x - 2)^{3/2} + \frac{5}{18} (3x - 2)^{1/2} + C$

Let me know if you need further explanation or help with other problems!