If `int (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

Options

A. `(2)/(3)(x+2)`
B. `(1)/(3)(x+4)`
C. `(2)/(3)(x-4)`
D. `(1)/(3)(x+1)`

Explanation

To solve the integral \( \int \frac{x+1}{\sqrt{2x-1}} \, dx \) and find \( f(x) \) such that

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

we will follow these steps:

### Step 1: Substitution
Let \( t = \sqrt{2x - 1} \). Then, we have:

\[ t^2 = 2x - 1 \implies 2x = t^2 + 1 \implies x = \frac{t^2 + 1}{2}. \]

### Step 2: Differentiate to find \( dx \)
Now, differentiate \( t \) with respect to \( x \):

\[ \frac{dt}{dx} = \frac{1}{2\sqrt{2x - 1}} \cdot 2 = \frac{1}{\sqrt{2x - 1}} \implies dx = \sqrt{2x - 1} \, dt = t \, dt. \]

### Step 3: Substitute in the integral
Now substitute \( x \) and \( dx \) in the integral:

\[ \int \frac{x + 1}{\sqrt{2x - 1}} \, dx = \int \frac{\frac{t^2 + 1}{2} + 1}{t} \cdot t \, dt. \]

This simplifies to:

\[ \int \left(\frac{t^2 + 1}{2} + 1\right) dt = \int \left(\frac{t^2 + 1 + 2}{2}\right) dt = \int \left(\frac{t^2 + 3}{2}\right) dt. \]

### Step 4: Integrate
Now, integrate:

\[ = \frac{1}{2} \int (t^2 + 3) \, dt = \frac{1}{2} \left(\frac{t^3}{3} + 3t\right) + C = \frac{t^3}{6} + \frac{3t}{2} + C. \]

### Step 5: Substitute back for \( t \)
Now substitute back \( t = \sqrt{2x - 1} \):

\[ = \frac{(\sqrt{2x - 1})^3}{6} + \frac{3\sqrt{2x - 1}}{2} + C = \frac{(2x - 1)^{3/2}}{6} + \frac{3(2x - 1)^{1/2}}{2} + C. \]

### Step 6: Factor out \( \sqrt{2x - 1} \)
Now, we can factor out \( \sqrt{2x - 1} \):

\[ = \sqrt{2x - 1} \left(\frac{(2x - 1)}{6} + \frac{3}{2}\right) + C. \]

### Step 7: Simplify the expression
Combine the terms inside the parentheses:

\[ = \sqrt{2x - 1} \left(\frac{2x - 1 + 9}{6}\right) + C = \sqrt{2x - 1} \left(\frac{2x + 8}{6}\right) + C. \]

### Step 8: Identify \( f(x) \)
Thus, we have:

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

where

\[ f(x) = \frac{2x + 8}{6} = \frac{x + 4}{3}. \]

### Final Answer
Therefore, the function \( f(x) \) is:

\[ \boxed{\frac{x + 4}{3}}. \]