∫ arctan(4t) d t {\displaystyle \int {\text{arctan(4t)}}dt}
u = arctan(4t) d v = d t {\displaystyle u={\text{arctan(4t)}}\qquad dv=dt}
d u = 4 1 + ( 4 t ) 2 d t v = t {\displaystyle du={\frac {4}{1+(4t)^{2}}}dt\qquad v=t}
∫ arctan(4t) d t = tarctan(4t) − 4 ∫ t 1 + 16 t 2 d t = π [ 4 x − x 2 + 1 12 x 3 ] | 1 2 = π [ ( 4 ( 2 ) − ( 2 ) 2 + 1 12 ( 2 ) 3 ) − ( 4 ( 1 ) − ( 1 ) 2 + 1 12 ( 1 ) 3 ) ] = π [ ( 8 − 4 + 8 12 ) − ( 4 − 1 + 1 12 ) ] = π [ 4 + 8 12 − 3 − 1 12 ] = π [ 1 + 7 12 ] = π [ 12 12 + 7 12 ] = π [ 19 12 ] = 19 π 12 {\displaystyle {\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\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}}}
![{\displaystyle {\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/41b8b7d449037af14758a6dfc20614834eea5c36)