What is ∫(e^ x + x) x d x equal to ?

$$\begin{aligned} & \int\left(e^{\log x}+\sin x\right) \cdot \cos x \cdot d x=\int(x+\sin x) \cdot \cos x \cdot d x \\ & =\int(x \cdot \cos x+\sin x \cdot \cos x) \cdot d x\end{aligned}$$$$\begin{aligned} & =\int x \cdot \cos x \cdot d x+\int \sin x \cdot \cos x \cdot d x \\ & =x \cdot \int \cos x \cdot d x-\int\left(\frac{d x}{d x} \cdot \int(\cos x) \cdot d x\right) \cdot d x \\ & \quad+\int \sin x \cdot \cos x \cdot d x\end{aligned}$$$$\begin{aligned} & =x \cdot \sin x-\int \sin x \cdot d x+\int \sin x \cdot \cos x \cdot d x \\ & =x \cdot \sin x-(-\cos x)+\int \sin x \cdot \cos x \cdot d x\end{aligned}$$=x \sin x+\cos x+\int \sin x \cdot \cos x \cdot d x$Now,$\int \sin x \cdot \cos x \cdot d x$Let$\sin x=z, d z=\cos x \cdot d x\int \sin x \cdot \cos x d x=\int z \cdot d z=\frac{z^2}{2}=\frac{\sin ^2 x}{2}\therefore I=x \cdot \sin x+\cos x+\frac{\sin ^2 x}{2}+C$