Cho hàm số $f(x)$ thỏa mãn ${f}'(x)=(x+1){{e}^{x}}$ và $f(0)=1$...

Ta có: $f(x)=\int{(x+1){{e}^{x}}d\text{x}}=\int{x.{{e}^{x}}d\text{x}}+\int{{{e}^{x}}d\text{x}}={{e}^{x}}+\int{x.{{e}^{x}}d\text{x}}$
Tính $I=\int{{{e}^{x}}d\text{x}}$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{x}}d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v={{e}^{x}} \\
\end{aligned} \right.$
$\Rightarrow I=\int{x{{e}^{x}}d\text{x}}=x{{e}^{x}}-\int{{{e}^{x}}d\text{x}}=x{{e}^{x}}-{{e}^{x}}+C$
$\Rightarrow f(x)={{e}^{x}}+x{{e}^{x}}-{{e}^{x}}+C=x{{e}^{x}}+C$
Lại có: $f(0)=1\Rightarrow 0.{{e}^{0}}+C=1\Rightarrow C=1$
$f(x)=x{{e}^{x}}+1\Rightarrow f(2)=2{{\text{e}}^{2}}+1=2{{\text{e}}^{2}}+1$.