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

Options

A. `2pi`
B. `pi-2`
C. `2-(pi)/(2)`
D. None of these

Answer

To solve the integral

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

we can utilize the property of definite integrals that states:

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

In our case, \( a = -1 \) and \( b = 1 \). Therefore, we can rewrite the integral as:

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

Now, substituting \( \sin(-x) = -\sin x \) and noting that \( (-x)^2 = x^2 \), we have:

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

Now, let's denote this integral as \( J \):

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

Now we can add the two integrals \( I \) and \( J \):

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

This simplifies to:

\[ I + J = \int_{-1}^{1} \frac{0}{1 + x^2} \, dx = 0. \]

Since \( J = -I \), we have:

\[ I - I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0. \]

Thus, the value of the integral is:

\[ \boxed{0}. \]