∫ 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 = tarctan(4t) − 4 32 ∫ 1 u = tarctan(4t) − 1 8 i n ( u ) = tarctan(4t) − 1 8 i n ( 1 + 16 t 2 ) + C {\displaystyle \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}}={\text{tarctan(4t)}}-{\frac {1}{8}}in(u)={\text{tarctan(4t)}}-{\frac {1}{8}}in(1+16t^{2})+C}
\begin{align} u=1+16t^{2} \\[1ex] du=32t dt \\[1ex] \frac{1}{32}du=t dt
\end{align} </math>
