The value of the integral `int_(-1)^(1){(x^(2013))/(e^(|x|)(x^(2)+cosx))+(1)/(e^(|x|))}dx` is equal to

Options

A. 0
B. `1-e^(-1)`
C. `2e^(-1)`
D. `2(1-e^(-1))`

Answer

To solve the integral

\[ I = \int_{-1}^{1} \left( \frac{x^{2013}}{e^{|x|}(x^2 + \cos x)} + \frac{1}{e^{|x|}} \right) dx, \]

we can break this integral into two parts and use the properties of definite integrals.

### Step 1: Split the Integral
We can express the integral as:

\[ I = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^2 + \cos x)} \, dx + \int_{-1}^{1} \frac{1}{e^{|x|}} \, dx. \]

Let’s denote these two integrals as \( I_1 \) and \( I_2 \):

\[ I_1 = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^2 + \cos x)} \, dx, \]
\[ I_2 = \int_{-1}^{1} \frac{1}{e^{|x|}} \, dx. \]

### Step 2: Evaluate \( I_1 \)
Notice that \( x^{2013} \) is an odd function, and \( e^{|x|} \) and \( x^2 + \cos x \) are even functions. Therefore, \( I_1 \) can be evaluated as follows:

\[ I_1 = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^2 + \cos x)} \, dx = 0. \]

This is because the integral of an odd function over a symmetric interval around zero is zero.

### Step 3: Evaluate \( I_2 \)
Now we evaluate \( I_2 \):

\[ I_2 = \int_{-1}^{1} \frac{1}{e^{|x|}} \, dx. \]

We can split this integral into two parts:

\[ I_2 = \int_{-1}^{0} \frac{1}{e^{-x}} \, dx + \int_{0}^{1} \frac{1}{e^{x}} \, dx. \]

### Step 4: Calculate Each Part of \( I_2 \)
1. For \( x \) in \([-1, 0]\):

\[ \int_{-1}^{0} e^{x} \, dx = [e^{x}]_{-1}^{0} = e^{0} - e^{-1} = 1 - \frac{1}{e}. \]

2. For \( x \) in \([0, 1]\):

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

### Step 5: Combine Results for \( I_2 \)
Now, we can combine both parts:

\[ I_2 = \left( 1 - \frac{1}{e} \right) + \left( 1 - \frac{1}{e} \right) = 2 - \frac{2}{e}. \]

### Step 6: Final Result
Since \( I_1 = 0 \), we have:

\[ I = I_1 + I_2 = 0 + \left( 2 - \frac{2}{e} \right) = 2 - \frac{2}{e}. \]

Thus, the value of the integral is:

\[ \boxed{2 - \frac{2}{e}}. \]