化简被积函数：
$$
\int \dfrac{1}{e^{3x}+e^{-3x}}\,dx=\int \dfrac{1}{e^{3x}+\dfrac{1}{e^{3x}}}\,dx=\int \dfrac{e^{3x}}{e^{6x}+1}\,dx.
$$
令 $u=e^{3x}$，则 $du=3e^{3x}dx$，即 $e^{3x}dx=\dfrac{1}{3}du$，代入得
$$
\int \dfrac{e^{3x}}{e^{6x}+1}\,dx=\int \dfrac{1}{u^2+1}\cdot\dfrac{1}{3}\,du=\dfrac{1}{3}\arctan u+C=\dfrac{1}{3}\arctan e^{3x}+C.
$$
故答案为 $\dfrac{1}{3}\arctan e^{3x}+C$。