Answer
$\tan (2x^2-1)+C$
Work Step by Step
Our aim is to solve the integral $ \int 4x \sec^2 (2x^2-1)\ dx$
Let us consider that $a =2x^2-1$ and $\dfrac{da}{dx}=4x \implies dx=\dfrac{1}{4x} \ da$
Now, $ nt 4x \sec^2 (2x^2-1)\ dx = \int 4x \sec^2 a \times \dfrac{1}{4x} \ da$
or, $= \int \sec^2 a \ da $
or, $=\tan a+C$
or, $=\tan (2x^2-1)+C$
