Cho F(x) là một nguyên hàm của hàm số  \(\begin{equation} f(x)=\frac{1}{2 \mathrm{e}^{x}+3} \text { thỏa mãn } F(0)=10 . \end{equation}\) Tìm F(x).

\(\begin{equation}
\begin{array}{l}
\text { Ta có } F(x)=\int f(x) \mathrm{d} x=\int \frac{1}{2 \mathrm{e}^{x}+3} \mathrm{~d} x \\
\text { Đặt } u=2 \mathrm{e}^{x}+3 \Rightarrow \mathrm{d} u=2 \mathrm{e}^{x} \mathrm{~d} x \Rightarrow \mathrm{d} x=\frac{\mathrm{d} u}{2 \mathrm{e}^{x}}=\frac{\mathrm{d} u}{u-3} . \\
\text { Khi đó } F(x)=\int \frac{1}{u(u-3)} \mathrm{d} u=\frac{1}{3}(\ln |u-3|-\ln |u|)+C=\frac{1}{3}\left(\ln \left(2 \mathrm{e}^{x}\right)-\ln \left(2 \mathrm{e}^{x}+3\right)\right)+C . \\
\text { Ta có } F(0)=10 \Leftrightarrow \frac{1}{3}(\ln 2-\ln 5)+C=10 \Leftrightarrow C=10-\frac{1}{3} \ln 2+\frac{1}{3} \ln 5 . \\
\text { Vậy } F(x)=\frac{1}{3}\left(\ln \left(2 \mathrm{e}^{x}\right)-\ln \left(2 \mathrm{e}^{x}+3\right)\right)+10-\frac{1}{3} \ln 2+\frac{1}{3} \ln 5=\frac{1}{3}\left(x-\ln \left(2 \mathrm{e}^{x}+3\right)\right)+10+\frac{\ln 5}{3} .
\end{array}
\end{equation}\)