To evaluate the integral
\[
I = \int_{-\pi}^{\pi} \frac{x \sin x}{e^x + 1} \, dx,
\]
we can utilize the property of definite integrals which states that
\[
\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx.
\]
In this case, \( a = -\pi \) and \( b = \pi \), so \( a + b = 0 \). Therefore, we can rewrite the integral as:
\[
I = \int_{-\pi}^{\pi} \frac{(-x) \sin(-x)}{e^{-x} + 1} \, dx.
\]
Since \( \sin(-x) = -\sin(x) \), we have:
\[
I = \int_{-\pi}^{\pi} \frac{-x (-\sin x)}{e^{-x} + 1} \, dx = \int_{-\pi}^{\pi} \frac{x \sin x}{e^{-x} + 1} \, dx.
\]
Now, we can denote this new integral as \( I_2 \):
\[
I_2 = \int_{-\pi}^{\pi} \frac{x \sin x}{e^{-x} + 1} \, dx.
\]
Next, we can add the two integrals \( I \) and \( I_2 \):
\[
2I = \int_{-\pi}^{\pi} \left( \frac{x \sin x}{e^x + 1} + \frac{x \sin x}{e^{-x} + 1} \right) \, dx.
\]
To combine the fractions, we find a common denominator:
\[
2I = \int_{-\pi}^{\pi} \frac{x \sin x (e^{-x} + 1) + x \sin x (e^x + 1)}{(e^x + 1)(e^{-x} + 1)} \, dx.
\]
This simplifies to:
\[
2I = \int_{-\pi}^{\pi} \frac{x \sin x (e^{-x} + e^x + 2)}{(e^x + 1)(e^{-x} + 1)} \, dx.
\]
Now, observe that \( e^{-x} + e^x = 2 \cosh x \) and \( (e^x + 1)(e^{-x} + 1) = e^x + e^{-x} + 2 = 2 \cosh x + 2 \).
Thus, we can rewrite \( 2I \) as:
\[
2I = \int_{-\pi}^{\pi} \frac{x \sin x (2 \cosh x + 2)}{2(\cosh x + 1)} \, dx.
\]
This simplifies to:
\[
I = \int_{-\pi}^{\pi} \frac{x \sin x}{\cosh x + 1} \, dx.
\]
Now, since \( x \sin x \) is an odd function (as \( x \) is odd and \( \sin x \) is odd), the integral of an odd function over a symmetric interval around zero is zero:
\[
I = 0.
\]
Thus, the final answer is:
\[
\int_{-\pi}^{\pi} \frac{x \sin x}{e^x + 1} \, dx = 0.
\]
