If $$\int \frac{x+1}{\sqrt{2x-1}} dx = f(x)\sqrt{2x-1} + C$$, where C is a constant of integration, then $$f(x)$$ is equal to:

We are given that

$$\int \frac{x+1}{\sqrt{2x-1}}\;dx = f(x)\sqrt{2x-1}+C,$$

where $$C$$ is the constant of integration. To identify $$f(x)$$, we differentiate the right-hand side and equate the derivative to the integrand, because differentiation is the inverse process of integration.

First, recall the product rule for differentiation: If $$u(x)$$ and $$v(x)$$ are differentiable, then $$\frac{d}{dx}\,[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$

Here we set $$u(x)=f(x), \qquad v(x)=\sqrt{2x-1}.$$ So

$$\frac{d}{dx}\,[f(x)\sqrt{2x-1}] = f'(x)\sqrt{2x-1} + f(x)\,\frac{d}{dx}\!\left(\sqrt{2x-1}\right).$$

Next we differentiate $$\sqrt{2x-1}$$. Using the chain rule, $$\frac{d}{dx}\!\left(\sqrt{2x-1}\right)=\frac{1}{2\sqrt{2x-1}}\cdot\frac{d}{dx}(2x-1)=\frac{1}{2\sqrt{2x-1}}\cdot 2=\frac{1}{\sqrt{2x-1}}.$$

Substituting this result back, we get

$$\frac{d}{dx}\,[f(x)\sqrt{2x-1}] = f'(x)\sqrt{2x-1} + f(x)\,\frac{1}{\sqrt{2x-1}}.$$

We combine the two terms over the common denominator $$\sqrt{2x-1}$$:

$$\frac{d}{dx}\,[f(x)\sqrt{2x-1}] = \frac{f'(x)(2x-1) + f(x)}{\sqrt{2x-1}}.$$

This derivative must equal the given integrand $$\dfrac{x+1}{\sqrt{2x-1}}$$. Therefore, we set the numerators equal:

$$f'(x)(2x-1) + f(x) = x + 1.$$

We now have the first-order linear differential equation

$$(2x-1)f'(x) + f(x) = x + 1.$$

The answer choices suggest that $$f(x)$$ is a linear function of the form $$ax + b$$. So we assume

$$f(x) = ax + b \quad\Longrightarrow\quad f'(x) = a.$$

Substituting these into the differential equation gives

$$(2x-1)\,a + (ax + b) = x + 1.$$

Expanding and collecting like terms, we obtain

$$2ax - a + ax + b = x + 1,$$ $$\bigl(2a + a\bigr)x + (b - a) = x + 1,$$ $$3a\,x + (b - a) = 1\,x + 1.$$

Equating coefficients of like powers of $$x$$, we get two equations:

Coefficient of $$x$$: $$3a = 1 \;\Longrightarrow\; a = \frac{1}{3},$$

Constant term: $$b - a = 1 \;\Longrightarrow\; b = 1 + a = 1 + \frac{1}{3} = \frac{4}{3}.$$

Thus

$$f(x) = \frac{1}{3}x + \frac{4}{3} = \frac{1}{3}(x + 4).$$

This matches Option D.

Hence, the correct answer is Option D.