Answer
$\dfrac{\sec (e^t-1) \space \tan (e^t-1)}{2}+\dfrac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$
Work Step by Step
Use the Reduction Formula:
$\int \sec^naxdx=\dfrac{\sec^{n-2}ax\tan ax}{a(n-1)}+\dfrac{n-2}{n-1}\int\sec^{(n-2)}(ax) \space dx$
Let $I=\int e^t\sec^3(e^t-1)dt$
Suppose $u=e^t-1 \implies du=e^tdt $
Therefore, $A=\int \sec^3(u) du\\=\dfrac{\sec u \space \tan u}{2}+\dfrac{1}{2}\int\sec (u) du \\=\dfrac{\sec (e^t-1) \space \tan (e^t-1)}{2}+\dfrac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$
