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

Options

A. increasing in `(0,oo)" and decreasing "(-oo,0)`
B. increasing in `(-oo,0)" and " decreasing in "(0,oo)`
C. increasing in `(-oo,oo)`
D. decreasing in `(-oo,oo)`

Answer

To solve the problem, we need to analyze the function given by the integral:

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

### Step 1: Differentiate \( f(x) \)

To determine whether \( f(x) \) is increasing or decreasing, we first compute the derivative \( f'(x) \):

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

### Step 2: Analyze the terms in \( f'(x) \)

Next, we need to analyze the expression \( f'(x) \):

1. The term \( 2 \) is a constant.
2. The term \( \frac{1}{1+x^2} \) is always positive and decreases as \( x \) increases.
3. The term \( \frac{1}{\sqrt{1+x^2}} \) is also always positive and decreases as \( x \) increases.

### Step 3: Check the bounds of the terms

We know that:

- \( \frac{1}{1+x^2} \leq 1 \) for all \( x \).
- \( \frac{1}{\sqrt{1+x^2}} \leq 1 \) for all \( x \).

Thus, both terms \( \frac{1}{1+x^2} \) and \( \frac{1}{\sqrt{1+x^2}} \) are less than or equal to 1. Therefore, we can conclude:

\[ f'(x) = 2 - \left( \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} \right) \geq 2 - 2 = 0 \]

### Step 4: Conclusion about \( f'(x) \)

Since \( f'(x) \geq 0 \) for all \( x \), we can conclude that \( f(x) \) is a non-decreasing function.

### Step 5: Determine if \( f(x) \) is strictly increasing

To check if \( f(x) \) is strictly increasing, we need to ensure that \( f'(x) \) is not equal to zero for any \( x \):

- At \( x = 0 \):
\[ f'(0) = 2 - 1 - 1 = 0 \]
- For \( x \neq 0 \), both \( \frac{1}{1+x^2} \) and \( \frac{1}{\sqrt{1+x^2}} \) are strictly less than 1, which implies that \( f'(x) > 0 \).

Thus, \( f(x) \) is strictly increasing for \( x \neq 0 \).

### Final Answer

The function \( f(x) \) is strictly increasing for all \( x \) except at \( x = 0 \) where it is constant.

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