Question

can someone help integrate this

Answer 1

\[\int\limits_{?}^{?} \frac{ (arctanx)^6 }{ 1+x^2 }\]

it's an indefinite integral so there is no interval (no "?")

Answer 2

i know there is some property that i can use but I'm not sure what it is

Answer 3

Remember your derivative of arctan? :)

Answer 4

secxtanx ?

Answer 5

no no silly, that's derivative of secx.

Derivative of arctan, \(\Large\rm \frac{d}{dx}\arctan x=\frac{1}{1+x^2}\)

Answer 6

Hmm interestinggg :O

Answer 7

ohh yaa

Answer 8

so u-sub?

Answer 9

\[\Large\rm \int\limits\limits\frac{ (\arctan x)^6 }{1+x^2}=\int\limits\limits(\arctan x)^6~\frac{1}{1+x^2}dx\]Yah u-sub sounds good! :)

Answer 10

ok so u=arctanx
du = 1+x^2dx

Answer 11

so integral of (u)^6 du ?

Answer 12

du=1/(1+x^2)dx

yes

Answer 13

gotta remember some of those tricky derivatives! :) hehe

Answer 14

got it thanks!

Answer 15

cool c: