7.1 Integration By Parts/11

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

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

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

${\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt\\[1ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[1ex]&=\pi \left[\left(4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}\right)-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[\left(8-4+{\frac {8}{12}}\right)-\left(4-1+{\frac {1}{12}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{12}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}$