Answer
$$\int e^t\sec^3(e^t-1)dt=\frac{\sec (e^t-1)\tan (e^t-1)}{2}+\frac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$$
Work Step by Step
$$A=\int e^t\sec^3(e^t-1)dt$$
Set $u=e^t-1$, which means $$du=e^tdt$$
Therefore, $$A=\int \sec^3udu$$
Use Reduction Formula 99, which states that
$$\int \sec^naxdx=\frac{\sec^{n-2}ax\tan ax}{a(n-1)}+\frac{n-2}{n-1}\int\sec^{n-2}axdx$$
for $n=3$ and $a=1$:
$$A=\frac{\sec u\tan u}{2}+\frac{1}{2}\int\sec udu$$ $$A=\frac{\sec u\tan u}{2}+\frac{1}{2}\ln|\sec u+\tan u|+C$$ $$A=\frac{\sec (e^t-1)\tan (e^t-1)}{2}+\frac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$$
