Evaluate the following integrals:

Assume $e^{3 x}+1=t$

$\Rightarrow \mathrm{d}\left(\mathrm{e}^{3 \mathrm{x}}+1\right)=\mathrm{dt}$

$\Rightarrow 3 \mathrm{e}^{3 \mathrm{x}}=\mathrm{dt}$

$\Rightarrow \mathrm{e}^{3 \mathrm{x}}=\frac{\mathrm{dt}}{3}$

Substituting $t$ and dt in the given equation we get

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

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

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

But $t=e^{3 x}+1$

$\therefore$ The above equation becomes

$\Rightarrow \frac{1}{3} \ln \left|e^{3 x}+1\right|+c$