If `int(1)/(x^(2)+2x+2)dx=f (x) +C` , then f (x)=

Options

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

Explanation

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

### Step 1: Rewrite the Denominator
First, we need to rewrite the denominator \( x^2 + 2x + 2 \) in a more manageable form. We can complete the square:

\[ x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1 \]

### Step 2: Substitute into the Integral
Now we can substitute this back into the integral:

\[ \int \frac{1}{(x + 1)^2 + 1} \, dx \]

### Step 3: Use the Formula for Inverse Tangent
The integral \( \int \frac{1}{x^2 + a^2} \, dx \) is known to be \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). In our case, \( a = 1 \) and we have \( (x + 1)^2 + 1 \). Thus, we can apply the formula:

\[ \int \frac{1}{(x + 1)^2 + 1} \, dx = \tan^{-1}(x + 1) + C \]

### Step 4: Write the Final Result
Thus, we have:

\[ f(x) = \tan^{-1}(x + 1) + C \]

### Conclusion
The final answer is:

\[ f(x) = \tan^{-1}(x + 1) \]