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Math Help - Integration with arctan

  1. #1
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    Integration with arctan

    \int x arctan (7x) dx

    I end up with \int x arctan (7x) = x / (7x^2+1) - \int 1 / (7x^2+1)

    I don't know how to evaluate the integral for that last one. Am I supposed to use x as my dv instead?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    \int x arctan (7x) dx

    I end up with \int x arctan (7x) = x / (7x^2+1) - \int 1 / (7x^2+1)

    I don't know how to evaluate the integral for that last one. Am I supposed to use x as my dv instead?
    u = \arctan(7x)

    dv = x \, dx<br />

    du = \frac{7}{1 + 49x^2} \, dx

    v = \frac{x^2}{2}

    \int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \int \frac{1}{2} \cdot \frac{7x^2}{1 + 49x^2} \, dx

    \int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{14} \int \frac{49x^2+1-1}{1 + 49x^2} \, dx

    \int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{14} \int 1 - \frac{1}{1 + 49x^2} \, dx

    \int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{98} \int 7 - \frac{7}{1 + 49x^2} \, dx

    \int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{x}{14} + \frac{\arctan(7x)}{98} + C

    \int x \arctan(7x) \, dx = \left(\frac{x^2}{2}+\frac{1}{98}\right)\arctan(7x) - \frac{x}{14} + C
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  3. #3
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    ^How did you manage to take out 1/14 and why did you add a negative one to the numerator?
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  4. #4
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    \int \frac{1}{2} \cdot \frac{7x^2}{1 + 49x^2} \, dx

    \frac{1}{2} \int \frac{7x^2}{1 + 49x^2} \, dx

    \frac{1}{2} \int \frac{7}{7} \cdot \frac{7x^2}{1 + 49x^2} \, dx

    \frac{1}{14} \int \frac{49x^2}{1 + 49x^2} \, dx

    \frac{1}{14} \int \frac{49x^2+1-1}{1 + 49x^2} \, dx

    \frac{1}{14} \int \frac{49x^2+1}{1 + 49x^2} - \frac{1}{1+49x^2} \, dx

    \frac{1}{14} \int 1 - \frac{1}{1+49x^2} \, dx
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