7.1 Integration By Parts/11

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

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

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

${\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt={\text{tarctan(4t)}}-{\frac {4}{32}}\int {\frac {1}{u}}\\[1ex]u=1+16t^{2}\\[1ex]du=32tdt\\[1ex]{\frac {1}{32}}du=tdt\end{aligned}}$

&= \text {tarctan(4t)} \\[1ex] &= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex] &= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex] &= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex] &= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex] &= \frac{19\pi}{12}

\end{align} </math>