If `f(x)=int_(x)^(x^(2))(dt)/((logt)^(2)),xne0` then `f(x)` is

Options

A. monotonically increasing in `(2,oo)`
B. monotonically decreasing in (1, 2)
C. monotonically decreasing in `(2,oo)`
D. monotonically decreasing in (0, 1)

Answer

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

\[ f(x) = \int_{x}^{x^2} \frac{dt}{(\log t)^2}, \quad x \neq 0 \]

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

To find the behavior of \( f(x) \), we will differentiate it using the Leibniz rule for differentiation under the integral sign:

\[ f'(x) = \frac{d}{dx} \left( \int_{x}^{x^2} \frac{dt}{(\log t)^2} \right) \]

According to the Leibniz rule, we have:

\[ f'(x) = \frac{d}{dx}(x^2) \cdot \frac{1}{(\log(x^2))^2} - \frac{d}{dx}(x) \cdot \frac{1}{(\log x)^2} \]

### Step 2: Calculate the derivatives

Calculating the derivatives:

1. The derivative of \( x^2 \) is \( 2x \).
2. The derivative of \( x \) is \( 1 \).

Thus, we can write:

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

### Step 3: Simplify \( f'(x) \)

Now we simplify \( f'(x) \):

Using the property of logarithms, \( \log(x^2) = 2 \log x \), we have:

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

This simplifies to:

\[ f'(x) = \frac{2x}{4 (\log x)^2} - \frac{1}{(\log x)^2} = \frac{x}{2 (\log x)^2} - \frac{1}{(\log x)^2} \]

Combining the terms gives:

\[ f'(x) = \frac{x - 2}{2 (\log x)^2} \]

### Step 4: Analyze the sign of \( f'(x) \)

To determine where \( f'(x) \) is positive or negative, we analyze:

\[ f'(x) = \frac{x - 2}{2 (\log x)^2} \]

1. The denominator \( 2(\log x)^2 \) is always positive for \( x > 1 \).
2. The numerator \( x - 2 \) is positive when \( x > 2 \).

Thus, \( f'(x) > 0 \) when \( x > 2 \).

### Conclusion

Since \( f'(x) > 0 \) for \( x > 2 \), we conclude that \( f(x) \) is monotonically increasing for \( x > 2 \).

### Final Answer

Thus, the correct option is that \( f(x) \) is monotonically increasing for \( x > 2 \).

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