Question

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

Answer 1

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

Answer 2

 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