Ask questions and get helpful responses.

Calculus 2

integral of xarctanxdx using integration by parts.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. ∫x arctan x dx

    u = arctan x
    du = 1/(1+x^2)

    dv = x dx
    v = 1/2 x^2

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

    now, x^2/(1+x^2) = 1 - 1/(1+x^2)

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

    so we wind up with

    1/2 (x^2 arctan(x) - (x - arctan(x)))
    = 1/2 ((1+x^2)arctan(x) - x) + C

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. ∫xarctan(x) dx

    Integration by parts:

    ∫vdu = vu - ∫udv

    Where:
    v = arctan(x) du = xdx
    dv = 1/(x^2 + 1)dx u = x^2/2

    ∫xarctan(x)dx = (x^2/2)arctan(x) - ∫(x^2)/2(x^2 + 1)dx

    Factor out constants:

    (1/2)x^2arctan(x) - (1/2)∫x^2/(x^2+1)dx

    Long division in the integrand:

    (1/2)x^2arctan(x) - (1/2)∫(1-(1/(x^2+1))dx

    Separate integrand:

    (1/2)x^2arctan(x) - (1/2)∫1dx -(1/2)∫1/(x^2+1)dx

    Integrate:

    (1/2)x^2arctan(x) - (x/2) - (1/2)arctan(x) + C

    Factor:

    (1/2)(x^2arctan(x) - x - arctan(x)) + C

    Complete:

    ∫xarctan(x)dx = (1/2)(x^2arctan(x)-x-arctan(x)) + C

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. thank you so much! I was getting stuck at the long division part.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Still need help? You can ask a new question.