Biết rằng $\Large \int\limits_{\text{0}}^{1}{x{{e}^{{{x}^{2}}+2}}\text

Biết rằng $\Large \int\limits_{\text{0}}^{1}{x{{e}^{{{x}^{2}}+2}}\text{d}x}=\dfrac{a}{2}\left( {{e}^{b}}-{{e}^{c}} \right)$ với $\Large a,b,c\in \mathbb{Z}$. Giá trị của $\Large a+b+c$ bằng

• A. $\Large 4$.
• B. $\Large 7$.
• C. $\Large 5$.
• D. $\Large 6$.

Ta có: $\Large \int\limits_{\text{0}}^{1}{x{{e}^{{{x}^{2}}+2}}\text{d}x}=\dfrac{1}{2}\int\limits_{\text{0}}^{1}{{{e}^{{{x}^{2}}+2}}\text{d}\left( {{x}^{2}}+2 \right)}=\dfrac{1}{2}{{e}^{{{x}^{2}}+2}}\bigg|_0^1=\dfrac{1}{2}\left( {{e}^{3}}-{{e}^{2}} \right).$

Nên $\Large a=1$, $\Large b=3$, $\Large c=2$.

Vậy $\Large a+b+c=6$.