$\left\{ \begin{align} & u=x+1 \\ & dv=\left( {{e}^{x}}-3 \right)dx \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align} & du=dx \\ & v=\int{\left( {{e}^{x}}-3 \right)dx=\left( {{e}^{x}}-3x \right)} \\ \end{align} \right.$

$\Rightarrow I=\left. \left( x+1 \right)\left( {{e}^{x}}-3x \right) \right|_{0}^{1}-\int_{0}^{1}{\left( {{e}^{x}}-3x \right)dx}$

$=\left. \left( x+1 \right)\left( {{e}^{x}}-3x \right) \right|_{0}^{1}-\left. \left( {{e}^{x}}-\frac{3}{2}{{x}^{2}} \right) \right|_{0}^{1}=e-\frac{9}{2}$