Evaluate the definite integral $\int_{0}^{1} \left(x e^{x} + \sin \frac{\pi x}{4}\right) dx$.

(D) Let $I = \int_{0}^{1} \left(x e^{x} + \sin \frac{\pi x}{4}\right) dx$.
Using the linearity property of integrals,we have:
$I = \int_{0}^{1} x e^{x} dx + \int_{0}^{1} \sin \frac{\pi x}{4} dx$.
For the first part,we use integration by parts: $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = e^{x} dx$. Then $du = dx$ and $v = e^{x}$.
$\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x}$.
For the second part:
$\int \sin \frac{\pi x}{4} dx = -\frac{4}{\pi} \cos \frac{\pi x}{4}$.
Combining these,the antiderivative $F(x)$ is:
$F(x) = x e^{x} - e^{x} - \frac{4}{\pi} \cos \frac{\pi x}{4}$.
Applying the limits from $0$ to $1$:
$I = F(1) - F(0) = \left(1 \cdot e^{1} - e^{1} - \frac{4}{\pi} \cos \frac{\pi}{4}\right) - \left(0 \cdot e^{0} - e^{0} - \frac{4}{\pi} \cos 0\right)$.
$I = \left(e - e - \frac{4}{\pi} \cdot \frac{1}{\sqrt{2}}\right) - \left(0 - 1 - \frac{4}{\pi} \cdot 1\right)$.
$I = -\frac{2\sqrt{2}}{\pi} + 1 + \frac{4}{\pi} = 1 + \frac{4 - 2\sqrt{2}}{\pi}$.