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24 June, 17:51

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

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  1. 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
  2. 24 June, 20:26
    0
    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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