Integral of `(1)/(sqrt(x^(2)+4))` w.r.t. `(x^(2)+3)` is

Options

A. `sqrt(x^(2)+4)+c`
B. `(1)/(sqrt(x^(2)+4))+c`
C. `2sqrt(x^(2)+4)+c`
D. `3sqrt(x^(2)+3)+c`

Answer

To solve the integral \( \int \frac{1}{\sqrt{x^2 + 4}} \, d(x^2 + 3) \), we will follow these steps:

### Step 1: Rewrite the Integral
We know that the differential \( d(x^2 + 3) \) can be expressed in terms of \( dx \):
\[ d(x^2 + 3) = 2x \, dx \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{\sqrt{x^2 + 4}} \, d(x^2 + 3) = \int \frac{1}{\sqrt{x^2 + 4}} \cdot 2x \, dx \]

### Step 2: Substitute for Simplicity
Next, we will use a substitution to simplify the integral. Let:
\[ t = x^2 + 4 \]
Then, the differential \( dt \) is:
\[ dt = 2x \, dx \]
Now we can substitute \( t \) into our integral:
\[ \int \frac{2x \, dx}{\sqrt{x^2 + 4}} = \int \frac{dt}{\sqrt{t}} \]

### Step 3: Integrate
The integral of \( \frac{1}{\sqrt{t}} \) is:
\[ \int t^{-1/2} \, dt = 2t^{1/2} + C \]
Thus, we have:
\[ \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C \]

### Step 4: Substitute Back
Now we substitute back for \( t \):
\[ t = x^2 + 4 \]
So, we have:
\[ 2\sqrt{t} = 2\sqrt{x^2 + 4} \]

### Final Answer
Putting everything together, the final result of the integral is:
\[ \int \frac{1}{\sqrt{x^2 + 4}} \, d(x^2 + 3) = 2\sqrt{x^2 + 4} + C \]