Question

please help me, i got final exam tomorrow
integrate : [(2+arctan 3x)^(1/2) / 1+9x^2

Answer 1

Let\[3x =\theta\]

Answer 2

let 3x=theta?

Answer 3

the bottom is the derivative of the top :)

Answer 4

well, the top innnards

Answer 5

May be you have an easier way.

Answer 6

trig sub it lol.. no really

u = 2+tan^-1(3x)

du = 3/(1+9x^2) dx

dx = (1+9x^2)/3

Answer 7

(1+9x^2) u^(1/2)
---------------
(1+9x^2) 3

(1/3) [S] u^(1/2) du

Answer 8

Good, my way was much more labor.

Answer 9

lol.... i like laborious :)