7.1 Integration By Parts/11

$\int {\text{arctan(4t)}}dt$

$u={\text{arctan(4t)}}\qquad dv=dt$

$du={\frac {4}{1+(4t)^{2}}}dt\qquad v=t$

Failed to parse (syntax error): {\displaystyle \int \text {arctan(4t)}dt = \text {t \cdot \arctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} = \text {tarctan(4t)}-\frac{1}{8}\ln(u)= \text {tarctan(4t)}-\frac{1}{8}\ln(1+16t^{2})+C }

${\begin{aligned}&u=1+16t^{2}\\[1ex]&du=32tdt\\[1ex]&{\frac {1}{32}}du=tdt\end{aligned}}$