Calculus (3rd Edition)

$$\int sech^2(1-2x) dx =-\frac{1}{2}\tanh (1-2x)+c.$$
Let $u=1-2x$, then $du=-2dx$ and hence $$\int sech^2(1-2x) dx =-\frac{1}{2}\int sech^2udu =-\frac{1}{2}\tanh u+c\\ =-\frac{1}{2}\tanh (1-2x)+c.$$