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 a property of definite integrals. This property states that:

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

In our case, we have \( a = -1 \) and \( b = 1 \), so \( a + b = 0 \). Thus, we can rewrite the integral as:

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

Since \( (-x)^2 = x^2 \), we can simplify this to:

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

Now we can express \( e^{-x} \) in terms of \( e^x \):

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

Now we have two expressions for \( I \):

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

Adding these two equations gives:

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

This simplifies to:

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

Now we can compute the integral:

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

This integral can be evaluated using the fact that

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

Thus, we have:

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

So, we find that:

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

Thus, the value of the integral is

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