Question

Integral:

Answer 1

\[\int\limits \arctan4tdt\]

Answer 2

any ideas?

Answer 3

maybe set u=4t and dv=arctan?

Answer 4

arctan is the name of a function, so i dont think you can do that ....

Answer 5

:/ I need to approach by integration by parts... but I'm a little confused with this one

Answer 6

let u=t

Answer 7

might have the uv parts backwards

u = arctan(4t)
dv = t

might be better

Answer 8

http://www.wolframalpha.com/input/?i=integrate+arctan%284t%29+dt

this would suggest a by parts setup

Answer 9

since arctan 4t is in the left term, and we are trying to integrate it ... id say its best to let that be u

v = t, so dv = 1

Answer 10

u = arctan(4t) v = t
du = dv = 1


tan u = 4t

sec^2u du = 4 dt

du = 4 dt/(tan^2u + 1) = 4dt/(16t^2+1)

Answer 11

so that simplifies it quit alot:\[\int tan^{-1}(4t)dt=t~tan^{-1}(4t)-\int\frac{4t}{16t^2+1}dt\]

Answer 12

how do you get 16t^2 + 1 right there?