Answer
$-\frac{1}{4} \ln|\cos (4x+1)|+c $
Work Step by Step
Let $ u=4x+1$ and hence $ du=4dx $, then we have
$$
\int \tan(4x+1)dx=\frac{1}{4}\int \tan u du\\
=\frac{1}{4}\int \frac{\sin u}{\cos u} du=-\frac{1}{4} \ln|\cos u|+c\\
=-\frac{1}{4} \ln|\cos (4x+1)|+c
$$
where we used the facts that $(\cos u)'=-\sin u $ and $\int \frac{u'}{u}=\ln u $.
