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