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

Answer

To solve the integral \(\int \frac{x+1}{\sqrt{2x-1}} \, dx\) and express it in the form \(f(x) \sqrt{2x-1} + C\), we will follow these steps:

### Step 1: Substitution
Let \(t = \sqrt{2x - 1}\). Then, squaring both sides gives:
\[ t^2 = 2x - 1 \implies 2x = t^2 + 1 \implies x = \frac{t^2 + 1}{2} \]
Now, we differentiate \(t\) with respect to \(x\):
\[ \frac{dt}{dx} = \frac{1}{\sqrt{2x - 1}} \cdot 2 \implies dx = \frac{t}{\sqrt{2}} dt \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \(t\):
\[ \int \frac{x+1}{\sqrt{2x-1}} \, dx = \int \frac{\left(\frac{t^2 + 1}{2} + 1\right)}{t} \cdot \frac{t}{\sqrt{2}} dt \]
This simplifies to:
\[ \int \frac{\left(\frac{t^2 + 3}{2}\right)}{\sqrt{2}} dt = \frac{1}{2\sqrt{2}} \int (t^2 + 3) dt \]

### Step 3: Integrate
Now we perform the integration:
\[ \frac{1}{2\sqrt{2}} \left( \frac{t^3}{3} + 3t \right) + C = \frac{1}{6\sqrt{2}} t^3 + \frac{3}{2\sqrt{2}} t + C \]

### Step 4: Substitute Back
Now substitute back \(t = \sqrt{2x - 1}\):
\[ = \frac{1}{6\sqrt{2}} (2x - 1)^{\frac{3}{2}} + \frac{3}{2\sqrt{2}} \sqrt{2x - 1} + C \]

### Step 5: Factor Out \(\sqrt{2x - 1}\)
We can express this in the required form:
\[ = \sqrt{2x - 1} \left( \frac{1}{6\sqrt{2}} (2x - 1) + \frac{3}{2\sqrt{2}} \right) + C \]
Now, simplifying the expression inside the parentheses:
\[ = \sqrt{2x - 1} \left( \frac{2x - 1 + 9}{6\sqrt{2}} \right) = \sqrt{2x - 1} \left( \frac{2x + 8}{6\sqrt{2}} \right) \]

### Step 6: Identify \(f(x)\)
From the expression above, we can identify:
\[ f(x) = \frac{2x + 8}{6\sqrt{2}} = \frac{x + 4}{3} \]

Thus, the final answer is:
\[ \boxed{\frac{x + 4}{3}} \]