Verified Answer :
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) = 1arctan(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 = xarctan(3x) - ∫ (x (3/(1+9x^2))) dx
= xarctan(3x) - ∫ (3x / (1 + 9x^2)) dx
take w = 1 + 9x^2
dw = w' dx
dw = 18x dx
dx = dw / 18x
so now you have
= xarctan(3x) - ∫ (3x)/w dw/18x
the x's cancel and you end up with
= xarctan(3x) - (1/6)∫(1/w)dw
= xarctan(3x) - (1/6) ln|w|
= xarctan(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) = 1arctan(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 = xarctan(3x) - ∫ (x (3/(1+9x^2))) dx
= xarctan(3x) - ∫ (3x / (1 + 9x^2)) dx
take w = 1 + 9x^2
dw = w' dx
dw = 18x dx
dx = dw / 18x
so now you have
= xarctan(3x) - ∫ (3x)/w dw/18x
the x's cancel and you end up with
= xarctan(3x) - (1/6)∫(1/w)dw
= xarctan(3x) - (1/6) ln|w|
= xarctan(3x) - (ln(1 + 9x^2))/6.......1 + 9x^2 is always positive
your answer:
x*arctan(3x) - (ln(1 + 9x^2))/6