`int(1)/(sqrt(x^(2)+2))d(x^(2)+1)` is equal to

Options

A. `2sqrt(x^(2)+2)+C`
B. `2((x^(2)+2)^(5/2))+C`
C. `(1)/((x^(2)+2)^(3//2))+C`
D. none of these

Explanation

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

### Step 1: Rewrite the integral
We start by recognizing that \( d(x^2 + 1) = 2x \, dx \). Thus, we can rewrite the integral as:
\[ \int \frac{1}{\sqrt{x^2 + 2}} \cdot 2x \, dx \]

### Step 2: Use substitution
Let \( t = x^2 + 2 \). Then, we differentiate \( t \) to find \( dt \):
\[ dt = 2x \, dx \quad \Rightarrow \quad 2x \, dx = dt \]
Now we can substitute \( t \) into the integral:
\[ \int \frac{1}{\sqrt{t}} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{\sqrt{t}} \, dt \) can be solved using the power rule for integration:
\[ \int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C \]

### Step 4: Substitute back
Now we substitute back \( t = x^2 + 2 \):
\[ 2\sqrt{t} + C = 2\sqrt{x^2 + 2} + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int \frac{1}{\sqrt{x^2 + 2}} \, d(x^2 + 1) = 2\sqrt{x^2 + 2} + C \]

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