Question

integral[(tan^(-1)x)/(1+x^2)]

Answer 1

put -1/x=t and calculate dx/dt

Answer 2

hey...hang on...try integration by parts...

Answer 3

make x = tan(t)

Answer 4

put (tan^(-1)x) = t
1/1+x^2 dx = dt
integratl (t dt) = t^2/2

Answer 5

uzmas on the right track :)

Answer 6

tan(t) = x ; x = tan-^1(x)

dt sec^2 = dx ; 1+tan^2(t) = sec^2

tan^-1(x) t sec^2(t) dt
-------- --> -------------- --> t dt
1 + x^2 sec^2(t)

Answer 7

[S] t dt = (t^2)/2 substitute back for t

tan^-1(x)
-------- is what it would be if you see thru the typos above lol
2

Answer 8

+C :)

Answer 9

[ tan^-1(x) ]^2
----------- comeon people......let me know when im wrong ;)
2

Answer 10

+C....ack!!

Answer 11

you r right, but after substitution we can differentiate both sides

Answer 12

d/dx(tan ^(-1)x) = 1/!+ x^2

Answer 13

oh 1/1 + x^2

Answer 14

tan^-1(x) --> 1/(1+x^2)

Answer 15

yes right

Answer 16

n directly we get Integral (tdt)

Answer 17

and the (...)^2 --> 2(....)

2(.....)/2 = (.......) = tan^-1(x)/(1+x^2) :) yay!!

Answer 18

??

Answer 19

lol..... it makes sense to me :)

Answer 20

:)

Answer 21

So what method was used? trig substitution?

Answer 22

I used trig sub...yeah

Answer 23

i tend to forget that dt != dx ... and that there is something I miss :)