Answer
$$\int\sec\theta\tan\theta\sqrt{1+\sec\theta}d\theta=\frac{2(1+\sec\theta)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int\sec\theta\tan\theta\sqrt{1+\sec\theta}d\theta$$
We set $a=1+\sec\theta$, which means $$da=\sec\theta\tan\theta d\theta$$
Therefore, $$A=\int \sqrt ada=\int a^{1/2}da$$ $$A=\frac{a^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2a^{3/2}}{3}+C$$ $$A=\frac{2(1+\sec\theta)^{3/2}}{3}+C$$
