The problem is to evaluate the indefinite integral: $\int \sqrt{3x-5} dx$

1. Problem Description

The problem is to evaluate the indefinite integral:

$\int \sqrt{3x-5} dx$

2. Solution Steps

Let $u = 3x - 5$.

Then $\frac{du}{dx} = 3$, which implies $dx = \frac{1}{3} du$.

Substituting $u$ and $dx$ into the integral, we have:

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

Using the power rule for integration:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$

We get:

$\frac{1}{3} \int u^{\frac{1}{2}} du = \frac{1}{3} \cdot \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{1}{3} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{9} u^{\frac{3}{2}} + C$

Now, substitute $u = 3x - 5$ back into the expression:

$\frac{2}{9} u^{\frac{3}{2}} + C = \frac{2}{9} (3x - 5)^{\frac{3}{2}} + C$

3. Final Answer

$\int \sqrt{3x-5} dx = \frac{2}{9}(3x-5)^{\frac{3}{2}} + C$