∫ arctan(4t) d t ⋅ {\displaystyle \int {\text{arctan(4t)}}dt\cdot }
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 = tarctan(4t) − 4 32 ∫ 1 u = tarctan(4t) − 1 8 ln ( u ) = tarctan(4t) − 1 8 ln ( 1 + 16 t 2 ) + C u = 1 + 16 t 2 d u = 32 t d t 1 32 d u = t d t {\displaystyle {\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]&={\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=32tdt\\[1ex]&\qquad {\frac {1}{32}}du=tdt\end{aligned}}}
![{\displaystyle {\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]&={\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=32tdt\\[1ex]&\qquad {\frac {1}{32}}du=tdt\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6fd3195bb92764abdb3129e7de0df42dc01ec8d0)