Answer
$\displaystyle \frac{1}{3}\tanh(3x)+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$\displaystyle \int{\rm sech}^{2}udu=\tanh u+C$
----
$\displaystyle \int \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}^{2}(3x)dx=\left[\begin{array}{lll}
u=3x & & \\
du=3dx & \Rightarrow & \frac{1}{3}du=dx
\end{array}\right]$
$=\displaystyle \frac{1}{3}\int \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}^{2}\underbrace{(3x)}_{u}\underbrace{(3dx)}_{du}$
$=\displaystyle \frac{1}{3}\tanh(3x)+C$
