$\int x^2\left[\sqrt{2}\sin \left(\frac{\pi }{4}+x\right)+e^x\right]dx=$

The integral expression is given as,

$I=\int x^2\left[\sqrt{2}\left(\sin \frac{\mathrm{\pi }}{4}+x\right)+e^x\right]dx$

$\Rightarrow I=\int x^2\left[\sqrt{2}\left(\sin \frac{\mathrm{\pi }}{4}\cos x+\sin x\cos \frac{\mathrm{\pi }}{4}\right)+e^x\right]dx$

$\Rightarrow I=\int x^2\left[\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right)+e^x\right]dx$

Further simplifying we get,

$\Rightarrow I=\int x^2\left[\cos x+\sin x+e^x\right]dx$

$\Rightarrow I=\int x^2(\cos x+\sin x)dx+\int x^2{e}^{x}dx$

As we know from integration by parts,

$\int uvdx=u\int vdx-\int \left\{\frac{du}{dx}·\int vdx\right\}dx$

$\Rightarrow I=x^2\int \left(\cos x+\sin x\right)dx-\int \left\{\frac{d}{dx}\left(x^2\right)\int \left(\cos x+\sin x\right)dx\right\}+x^2{e}^x-\int 2x{e}^{x}dx$

$\Rightarrow I=x^2\left(\sin x-\cos x\right)-\int 2x\left(\sin x-\cos x\right)dx+x^2{e}^x-2\left[x{e}^x-e^x\right] +C$

$\left\{\because \int \sin xdx=-\cos x, \int \cos xdx=\sin x\right\}$

$\Rightarrow x^2\left(\sin x-\cos x\right)-\left[2x\left(-\cos x-\sin x\right)-2\int \left(-\cos x-\sin x\right)dx\right]+\left(x^2-2x+2\right)e^x+C$

$\Rightarrow x^2\left(\sin x-\cos x\right)+2x\left(\cos x+\sin x\right)-2\left(\sin x-\cos x\right)+\left(x^2-2x+2\right)e^x+C$

$\Rightarrow I=\left(x^2+2x-2\right)\sin x+\left(-x^2+2x+2\right)\cos x+\left(x^2-2x+2\right)e^x+C$