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Let $I=\int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x$

$\int\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x=x \int e^{x} d x-\int\left\{\left(\frac{d}{d x} x\right) \int e^{x} d x\right\} d x+\left\{\frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}}\right\}$

$=x e^{x}-\int e^{x} d x-\frac{4 \pi}{\pi} \cos \frac{x}{4}$

$=x e^{x}-e^{x}-\frac{4 \pi}{\pi} \cos \frac{x}{4}$

$=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

$I=\mathrm{F}(1)-\mathrm{F}(0)$

$=\left(1 . e^{1}-e^{1}-\frac{4}{\pi} \cos \frac{\pi}{4}\right)-\left(0 . e^{0}-e^{0}-\frac{4}{\pi} \cos 0\right)$

$=e-e-\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}\right)+1+\frac{4}{\pi}$

$=1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}$