Answer
$\tanh (x-\dfrac{1}{2}) +C$
Work Step by Step
As we are given that $\int sech^2(x-\dfrac{1}{2}) dx$
Now, plug $(x-\dfrac{1}{2})=t \implies dx =dt$
So, $\int sech^2(x-\dfrac{1}{2}) dx=\int (sech^2 t) dt=\tanh t +C= \tanh (x-\dfrac{1}{2}) +C$
