Integral of arctan(8x) dx.

Question

hartnn · Answer

put 8x=u
then dx= (1/8)du ok?

anonymous · Answer

Alright.

hartnn · Answer

so your new integral is
$$(1/8)\int\limits_{}^{}\tan^{-1}u \: du$$
right ?

anonymous · Answer

Indeed it is.

hartnn · Answer

now u know the integral of tan^-1 x ??

anonymous · Answer

I actually don't have it memorized xD I just started integration by parts so we're doing hw problems to show the proofs for them..

so i'm assuming:
dv = du
v= u
u= arctan(u)
du= 1/1+u^2

hartnn · Answer

that is correct except, v=1

anonymous · Answer

i know it equals though:

xarctan(x)-1/2ln(1+x^2)

anonymous · Answer

1, yea typo xD and i know what it is, but for some reason when i do the problem i don't get what it should.

hartnn · Answer

ok,what u got? 
how did u integrate u/(1+u^2) ?

hartnn · Answer

put another variable say y=1+u^2
dy = 2u du 
ok?

anonymous · Answer

yea, this is weird cuz usually i use u, but we kinda already used u-sub in the beginning xD but y works!

hartnn · Answer

ok,go ahead and ask if u still don't get it.

anonymous · Answer

alright i'll let ya know

anonymous · Answer

haha I got it xD
i know what i was doing wrong too xD i kept writing ln(1+8x^2) instead of 64x^2 xD i'm a dunce.. thanks again hartnn :)

hartnn · Answer

welcome :)