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

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi`
D. 0

Answer

To solve the integral

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

we can use the property of definite integrals which states that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx. \]

In our case, \( a = -1 \) and \( b = 1 \), so \( a + b = 0 \). Therefore, we can write:

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

Since \( (-x)^2 = x^2 \), we have:

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

Now, we can express \( 1 + e^{-x} \) as:

\[ 1 + e^{-x} = \frac{e^x + 1}{e^x}. \]

Thus, we can rewrite the integral as:

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

Now, we can add the two expressions for \( I \):

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

Combining the fractions gives:

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

Now we can evaluate the integral:

\[ \int_{-1}^{1} \frac{1}{1+x^2} \, dx. \]

This integral represents the area under the curve of the function \( \frac{1}{1+x^2} \) from -1 to 1, which is known to be:

\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x). \]

Evaluating this from -1 to 1 gives:

\[ \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}. \]

Thus, we have:

\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}. \]

Therefore, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]