$V=\pi \int_{0}^{1}{{{\left( \sqrt{x\left( {{e}^{x}}+1 \right)} \right)}^{2}}dx=\pi \int_{0}^{1}{x\left( {{e}^{x}}+1 \right)dx=\pi \int_{0}^{1}{x{{e}^{x}}dx+\left. \pi \frac{{{x}^{2}}}{2} \right|_{0}^{1}=\pi \int_{0}^{1}{x{{e}^{x}}dx+\frac{\pi }{2}}}}}$

+) Đặt $\left\{ \begin{align} & u=v \\ & dv={{e}^{x}}dx \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align} & du=dx \\  & v={{e}^{x}} \\ \end{align} \right.\Rightarrow \int_{0}^{1}{x{{e}^{x}}dx=\left. x{{e}^{x}} \right|_{0}^{1}-\int_{0}^{1}{{{e}^{x}}dx=e-\left. {{e}^{x}} \right|_{0}^{1}=1}}$

Do đó: $V'=\pi +\frac{\pi }{2}=\frac{3\pi }{2}(dvtt)$

Thật ra để tính $V=\pi \int_{0}^{1}{{{\left( \sqrt{x\left( {{e}^{x}}+1 \right)} \right)}^{2}}}dx$ ta dùng MTCT và dễ dàng ra đáp án B.