How to integrate arctan 3x. dx ... ?

To integrate by parts u need to take 2 functions, in this case, you only have arctan (3x) and therefore you second function will just be 1. this sounds confusing but think about it: arctan (3x) = 1*arctan (3x) ill take u = arctan (3x) dv = 1 - so you know du = 3 / (1+9x^2) ... derivative of tangent inverse v = x ∫ u dv = uv - ∫ v du ∫ (arctan (3x) * 1) dx = x*arctan (3x) - ∫ (x * (3 / (1+9x^2))) dx = x*arctan (3x) - ∫ (3x / (1 + 9x^2)) dx take w = 1 + 9x^2 dw = w' dx dw = 18x dx dx = dw / 18x so now you have = x*arctan (3x) - ∫ (3x) / w * dw/18x the x's cancel and you end up with = x*arctan (3x) - (1/6) * ∫ (1/w) dw = x*arctan (3x) - (1/6) * ln|w| = x*arctan (3x) - (ln (1 + 9x^2)) / 6 ... 1 + 9x^2 is always positive your answer: x*arctan (3x) - (ln (1 + 9x^2)) / 6 good luck

to integrate by parts u need to take 2 functions, in this case, you only have arctan (3x) and therefore you second function will just be 1. this sounds confusing but think about it:

arctan (3x) = 1*arctan (3x)

ill take

u = arctan (3x)

dv = 1

-so you know

du = 3 / (1+9x^2) ... derivative of tangent inverse

v = x

∫ u dv = uv - ∫ v du

∫ (arctan (3x) * 1) dx = x*arctan (3x) - ∫ (x * (3 / (1+9x^2))) dx

= x*arctan (3x) - ∫ (3x / (1 + 9x^2)) dx

take w = 1 + 9x^2

dw = w' dx

dw = 18x dx

dx = dw / 18x

so now you have

= x*arctan (3x) - ∫ (3x) / w * dw/18x

the x's cancel and you end up with

= x*arctan (3x) - (1/6) * ∫ (1/w) dw

= x*arctan (3x) - (1/6) * ln|w|

= x*arctan (3x) - (ln (1 + 9x^2)) / 6 ... 1 + 9x^2 is always positive

your answer:

x*arctan (3x) - (ln (1 + 9x^2)) / 6