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 \(S=7\).