Question

let, I=∫tan^-1x dx
=tan^-1x ∫1 dx-∫{d/dx tan^-1x ∫1 dx} dx
=x tan^-1x -∫x/1+x^2 dx
=x tan^-1x- ????

Answer 1

@amistre64 help

Answer 2

@dumbcow

Answer 3

@zepdrix

Answer 4

@dumbcow :(

Answer 5

ok trying to follow your work lol

looks like you did integration by parts and you are stuck at integrating x/(1+x^2)

use substitution:

w = 1 + x^2
dw = 2x dx ---> dx = dw/2x

"the x's will cancel"

Answer 6

so x tan^-1x -∫dz/2/z
=x tan^-1x-1/2∫1/z dz
=x tan^-1x-1/2 ln z
=x tan^-1x-1/2 ln (1+x^2) ans???

Answer 7

@zepdrix

Answer 8

help bro :'(

Answer 9

@dumbcow

Answer 10

yep :)

use this to check answers
http://www.wolframalpha.com/input/?i=integrate+arctan%28x%29+dx

Answer 11

but if i use the method of substitution in integration by parts then what if the teacher is gonna cut my number :(

Answer 12

thank you i will definitely take help from that :) thank you very very much :) bro

Answer 13

the teacher shouldn't....many integrals require multiple methods to finish

the substitution is done within the integration by parts so i think you are ok

Answer 14

thank you :)