Answer
$$-\frac{1}{4}\ln(\cos (4x+1))+c$$
Work Step by Step
Let $u=4x+1$, then $du =4dx$. Now, we have
$$\int \tan(4x+1)dx=\frac{1}{4}\int \tan udu\\
=\frac{1}{4}\int\frac{\sin u}{\cos u}du=-\frac{1}{4}\ln(\cos u)+c\\
=-\frac{1}{4}\ln(\cos (4x+1))+c.$$
