Question

integrate by parts: ∫arctan4tdt

Answer 1

\[u=\tan^{-1}(4t)\]\[dv=tdt\]

Answer 2

i think you saw an extra t in there

dv should just be dt

Answer 3

yes I did :)

Answer 4

\[u=\tan^{-1}(4t)\]\[dv=dt\]I'm tired now, g'night!

Answer 5

\[du = \frac{ 4 }{ 1+t^2 }dt\]
\[v = t\]

Answer 6

\[I = t*\arctan(4t) - 4\int\limits_{}^{}\frac{ t }{ 1+t^2 }dt\]

u-sub from here should do it