`int_(-a)^(a) log (x+ sqrt(x^(2) + 1 ) ) dx `

Options

A. `2 log (a^(2) +1)`
B. `2 log ( sqrt(a^(2) +1) -a)`
C. `0`
D. `2 log ( a+ sqrt(a^(2) + 1) )`

Answer

To solve the integral \( I = \int_{-a}^{a} \log(x + \sqrt{x^2 + 1}) \, dx \), we will use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx \]

### Step 1: Define the function
Let \( f(x) = \log(x + \sqrt{x^2 + 1}) \).

### Step 2: Find \( f(-x) \)
Now, we need to compute \( f(-x) \):

\[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}) \]

### Step 3: Combine \( f(x) \) and \( f(-x) \)
Next, we will find \( f(x) + f(-x) \):

\[ f(x) + f(-x) = \log(x + \sqrt{x^2 + 1}) + \log(-x + \sqrt{x^2 + 1}) \]

Using the logarithmic property \( \log a + \log b = \log(ab) \):

\[ f(x) + f(-x) = \log\left((x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1})\right) \]

### Step 4: Simplify the expression
Now, we simplify the product inside the logarithm:

\[ (x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1}) = (\sqrt{x^2 + 1})^2 - x^2 = x^2 + 1 - x^2 = 1 \]

Thus, we have:

\[ f(x) + f(-x) = \log(1) = 0 \]

### Step 5: Evaluate the integral
Now we can substitute back into the integral:

\[ I = \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx = \int_{0}^{a} 0 \, dx = 0 \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{-a}^{a} \log(x + \sqrt{x^2 + 1}) \, dx = 0 \]