The value of the definite integral `int e^(-x^4) (2+ln(x+sqrt(x^2+1))+5x^3-8x^4)dx` is equal to

Options

A. 4e
B. `4/e`
C. `2e`
D. `2/e`

Answer

To solve the definite integral

\[ I = \int_{-1}^{1} e^{-x^4} \left( 2 + \ln(x + \sqrt{x^2 + 1}) + 5x^3 - 8x^4 \right) dx, \]

we can break it down into simpler parts.

### Step 1: Break Down the Integral
We can express \(I\) as the sum of several integrals:

\[ I = \int_{-1}^{1} e^{-x^4} \cdot 2 \, dx + \int_{-1}^{1} e^{-x^4} \ln(x + \sqrt{x^2 + 1}) \, dx + \int_{-1}^{1} 5x^3 e^{-x^4} \, dx - \int_{-1}^{1} 8x^4 e^{-x^4} \, dx. \]

### Step 2: Analyze the Functions
1. **First Integral**: The function \(e^{-x^4}\) is even, and the constant \(2\) is also even. Therefore, the first integral is non-zero.

2. **Second Integral**: We need to check if \(\ln(x + \sqrt{x^2 + 1})\) is odd.
- For \(f(x) = \ln(x + \sqrt{x^2 + 1})\), we find \(f(-x) = \ln(-x + \sqrt{x^2 + 1})\).
- It can be shown that \(f(-x) = -f(x)\), confirming that this function is odd.

3. **Third Integral**: The term \(5x^3\) is odd, hence the integral over a symmetric interval around zero will be zero.

4. **Fourth Integral**: The term \(8x^4\) is even, and thus this integral will not be zero.

### Step 3: Evaluate the Non-Zero Integrals
Now we can simplify \(I\):

\[ I = \int_{-1}^{1} 2 e^{-x^4} \, dx - \int_{-1}^{1} 8x^4 e^{-x^4} \, dx. \]

### Step 4: Use Symmetry
Using the property of definite integrals for even functions, we can rewrite:

\[ I = 2 \int_{0}^{1} 2 e^{-x^4} \, dx - 2 \int_{0}^{1} 8x^4 e^{-x^4} \, dx. \]

This simplifies to:

\[ I = 4 \int_{0}^{1} e^{-x^4} \, dx - 16 \int_{0}^{1} x^4 e^{-x^4} \, dx. \]

### Step 5: Substitution for the Second Integral
Let \(u = x^4\), then \(du = 4x^3 dx\) or \(dx = \frac{du}{4x^3}\). The limits change from \(0\) to \(1\) in \(x\) to \(0\) to \(1\) in \(u\).

Thus, the second integral becomes:

\[ \int_{0}^{1} x^4 e^{-x^4} \, dx = \frac{1}{4} \int_{0}^{1} e^{-u} \, du = \frac{1}{4} \left[-e^{-u}\right]_{0}^{1} = \frac{1}{4}(1 - e^{-1}). \]

### Step 6: Substitute Back
Now substituting back into our expression for \(I\):

\[ I = 4 \int_{0}^{1} e^{-x^4} \, dx - 16 \cdot \frac{1}{4}(1 - e^{-1}) = 4 \int_{0}^{1} e^{-x^4} \, dx - 4(1 - e^{-1}). \]

### Step 7: Final Evaluation
The integral \(\int_{0}^{1} e^{-x^4} \, dx\) is a known integral, but for the purposes of this problem, we can leave it as is.

Thus, the final expression for \(I\) is:

\[ I = 4 \int_{0}^{1} e^{-x^4} \, dx - 4 + 4e^{-1}. \]

### Conclusion
The value of the definite integral is given by:

\[ I = 4 \left( \int_{0}^{1} e^{-x^4} \, dx + e^{-1} - 1 \right). \]