Biết $\int\limits_{0}^{1}{\frac{\pi {{x}^{3}}+{{2}^{x}}+\text{e}{{x}^{3}}{{.2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{m}+\frac{1}{\text{e}\ln n}\ln \left( p+\frac{\text{e}}{\text{e}+\pi } \right)$ với $m$, $n$, $p$ là các số nguyên dương. Tính tổng $S=m+n+p$.

Ta có $\int\limits_{0}^{1}{\frac{\pi {{x}^{3}}+{{2}^{x}}+\text{e}{{x}^{3}}{{.2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{3}}+\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}} \right)\text{d}x}$ $=\left. \frac{{{x}^{4}}}{4} \right|_{0}^{1}+\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+J.$

Tính $J=\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}$.

Đặt $\pi +\text{e}{{.2}^{x}}=t\Rightarrow \text{e}{{.2}^{x}}\ln 2\text{d}x=\text{d}t\Leftrightarrow {{2}^{x}}\text{d}x=\frac{1}{\text{e}.\ln 2}\text{d}t$.

Đổi cận: Khi $x=0$ thì $t=\pi +\text{e}$; khi $x=1$ thì $t=\pi +2\text{e}$.

Khi đó $J=\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{\text{e}\ln 2}\int\limits_{\pi +\text{e}}^{\pi +2\text{e}}{\frac{1}{t}\text{d}t}=\frac{1}{\text{e}\ln 2}\left. \ln \left| t \right| \right|_{\pi +\text{e}}^{\pi +2\text{e}}=\frac{1}{\text{e}\ln 2}\ln \left( 1+\frac{\text{e}}{\text{e}+\pi } \right)$.

Suy ra $\int\limits_{0}^{1}{\frac{\pi {{x}^{3}}+{{2}^{x}}+\text{e}{{x}^{3}}{{.2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+\frac{1}{\text{e}\ln 2}\ln \left( 1+\frac{\text{e}}{\text{e}+\pi } \right)$$\Rightarrow m=4$, $n=2$, $p=1$.

Vậy $S=7$.