## Calculus (3rd Edition)

$-\frac{1}{4} \ln|\cos (4x+1)|+c$
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 $\int \frac{u'}{u}=\ln u$.