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Salazar
24 June, 17:51
How to integrate arctan 3x. dx ... ?
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Adalyn Noble
24 June, 18:09
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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
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Admiral
24 June, 20:26
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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
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