Integrate with respect to `x:1/sqrt(2x+1)`

Options

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

Answer

To solve the integral \(\int \frac{1}{\sqrt{2x + 1}} \, dx\), we will follow these steps:

### Step 1: Substitution
Let:
\[ u = 2x + 1 \]
Then, differentiate \(u\) with respect to \(x\):
\[ \frac{du}{dx} = 2 \quad \Rightarrow \quad du = 2 \, dx \quad \Rightarrow \quad dx = \frac{du}{2} \]

### Step 2: Rewrite the Integral
Now, substitute \(u\) and \(dx\) into the integral:
\[ \int \frac{1}{\sqrt{2x + 1}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} \]
This simplifies to:
\[ \frac{1}{2} \int u^{-1/2} \, du \]

### Step 3: Integrate
Now, integrate \(u^{-1/2}\):
\[ \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot \left( \frac{u^{1/2}}{1/2} \right) + C = u^{1/2} + C \]

### Step 4: Substitute Back
Now substitute back \(u = 2x + 1\):
\[ u^{1/2} + C = \sqrt{2x + 1} + C \]

### Final Answer
Thus, the integral \(\int \frac{1}{\sqrt{2x + 1}} \, dx\) evaluates to:
\[ \sqrt{2x + 1} + C \]

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