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 (unknown function "\begin{align}"): {\displaystyle \begin{align} \int \text {arctan(4t)}dt = \text {t.\cdotarctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} \\[1ex] &= \text {tarctan(4t)}-\frac{1}{8}\ln(u)= \text {tarctan(4t)}-\frac{1}{8}\ln(1+16t^{2})+C \\[1ex] & \qquad u=1+16t^{2} \\[1ex] & \qquad du=32t dt \\[1ex] & \qquad \frac{1}{32}du=t dt \end{align} }