Evaluate the following integrals:

Assume $3 \sec x+5=t$

$d(3 \sec x+5)=d t$

$3 \sec x \tan x=d t$

$\operatorname{Sec} x \tan x=\frac{\mathrm{dt}}{3}$

Substitute $t$ and $d t$

We get

$\Rightarrow \frac{1}{3} \int \frac{\mathrm{dt}}{\mathrm{t}}$

$\Rightarrow \frac{1}{3} \ln |\mathrm{t}|+\mathrm{c}$

But $t=3 \sec x+5$

$\therefore$ the equation becomes

$\Rightarrow \frac{1}{3} \ln |3 \sec x+5|+c$