Lời giải chi tiết:

Chọn khai triển: \({{\left( 1+x \right)}^{2n}}=C_{2n}^{0}+C_{2n}^{1}.x+C_{2n}^{2}.{{x}^{2}}+C_{2n}^{3}.{{x}^{3}}+...+C_{2n}^{2n}.{{x}^{2n}}\)

Lấy tích phân hai vế ta được:

\(\begin{align}  & \int{{{\left( 1+x \right)}^{2n}}}dx=\int{\left( C_{2n}^{0}+C_{2n}^{1}.x+C_{2n}^{2}.{{x}^{2}}+C_{2n}^{3}.{{x}^{3}}+...+C_{2n}^{2n}.{{x}^{2n}} \right)}dx \\ & \Leftrightarrow \frac{{{\left( 1+x \right)}^{2n+1}}}{2n+1}=C_{2n}^{0}x+C_{2n}^{1}.\frac{{{x}^{2}}}{2}+C_{2n}^{2}.\frac{{{x}^{3}}}{3}+C_{2n}^{3}.\frac{{{x}^{4}}}{4}+...+C_{2n}^{2n}.\frac{{{x}^{2n+1}}}{2n+1}\,\,\ \ \ \ \ \left( * \right) \\\end{align}\)

Với \(x=1\) ta có: \(\left( * \right)\Leftrightarrow \frac{{{2}^{2n+1}}}{2n+1}=C_{2n}^{0}+\frac{C_{2n}^{1}}{2}+\frac{C_{2n}^{2}}{3}+\frac{C_{2n}^{3}}{4}+\frac{C_{2n}^{4}}{5}+.......+\frac{C_{2n}^{2n}}{2n+1}.\ \ \ \ \left( 1 \right)\)

Với \(x=-1\) ta có: \(\left( * \right)\Leftrightarrow 0=-C_{2n}^{0}+\frac{C_{2n}^{1}}{2}-\frac{C_{2n}^{2}}{3}+\frac{C_{2n}^{3}}{4}-\frac{C_{2n}^{4}}{5}+.......-\frac{C_{2n}^{2n}}{2n+1}.\ \ \ \left( 2 \right)\)

Lấy \(\left( 1 \right)-\left( 2 \right)\) ta được: \(\frac{{{2}^{2n+1}}}{2n+1}=2\left( C_{2n}^{0}+\frac{C_{2n}^{2}}{3}+\frac{C_{2n}^{4}}{5}+.....+\frac{C_{2n}^{2n}}{2n+1} \right)\)

\(\begin{array}{l}
\Leftrightarrow C_{2n}^0 + \frac{{C_{2n}^2}}{3} + \frac{{C_{2n}^4}}{5} + ..... + \frac{{C_{2n}^{2n}}}{{2n + 1}} = \frac{{{2^{2n + 1}}}}{{2n + 1}}.\frac{1}{2}\\
\Leftrightarrow C_{2n}^0 + \frac{{C_{2n}^2}}{3} + \frac{{C_{2n}^4}}{5} + ..... + \frac{{C_{2n}^{2n}}}{{2n + 1}} = \frac{{{2^{2n}}}}{{2n + 1}}\\
\Leftrightarrow \frac{{8192}}{{15}} = \frac{{{2^{2n}}}}{{2n + 1}}\\
\Leftrightarrow \left\{ \begin{array}{l}
2n + 1 = 15\\
{2^{2n}} = 8192
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
n = 7\\
n = \frac{1}{2}{\log _2}8192 = 6,5
\end{array} \right. \Rightarrow n \in \emptyset .
\end{array}\)

Chọn D.