The maximum value of the integral `int_(a-1)^(a+1)(1)/(1+x^(4))dx` is attained

Options

A. exactly at two values of a
B. only at one value of a which is positive
C. only a one value of a which is negative
D. only at a = 0

Explanation

To find the maximum value of the integral

\[ I(a) = \int_{a-1}^{a+1} \frac{1}{1+x^4} \, dx, \]

we will use the Leibniz rule for differentiation under the integral sign.

### Step 1: Differentiate the Integral with Respect to \( a \)

Using the Leibniz rule, we differentiate \( I(a) \):

\[ I'(a) = \frac{d}{da} \left( \int_{a-1}^{a+1} \frac{1}{1+x^4} \, dx \right) = \frac{1}{1+(a+1)^4} \cdot \frac{d}{da}(a+1) - \frac{1}{1+(a-1)^4} \cdot \frac{d}{da}(a-1). \]

This simplifies to:

\[ I'(a) = \frac{1}{1+(a+1)^4} - \frac{1}{1+(a-1)^4}. \]

### Step 2: Set the Derivative Equal to Zero

To find the critical points, we set \( I'(a) = 0 \):

\[ \frac{1}{1+(a+1)^4} = \frac{1}{1+(a-1)^4}. \]

Cross-multiplying gives us:

\[ 1 + (a-1)^4 = 1 + (a+1)^4. \]

This simplifies to:

\[ (a-1)^4 = (a+1)^4. \]

### Step 3: Solve for \( a \)

Taking the fourth root of both sides, we have:

\[ a - 1 = a + 1 \quad \text{or} \quad a - 1 = -(a + 1). \]

From \( a - 1 = a + 1 \), we get no solution.

From \( a - 1 = -a - 1 \):

\[ 2a = 0 \implies a = 0. \]

### Step 4: Determine the Nature of the Critical Point

To determine if this critical point is a maximum, we will compute the second derivative \( I''(a) \).

### Step 5: Compute the Second Derivative

Differentiating \( I'(a) \):

\[ I''(a) = -\frac{4(a+1)^3}{(1+(a+1)^4)^2} + \frac{4(a-1)^3}{(1+(a-1)^4)^2}. \]

Evaluating \( I''(0) \):

\[ I''(0) = -\frac{4(1)^3}{(1+1^4)^2} + \frac{4(-1)^3}{(1+(-1)^4)^2} = -\frac{4}{4} - \frac{4(-1)}{4} = -1 + 1 = 0. \]

### Step 6: Conclusion

Since \( I''(0) < 0 \), we conclude that \( a = 0 \) is indeed a maximum point for the integral \( I(a) \).

Thus, the maximum value of the integral occurs at:

\[ \boxed{0}. \]