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Math Help - How to integrate when the inverse tangent is present?

  1. #1
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    Smile How to integrate when the inverse tangent is present?

    Ok, one of my homework problems for my calculus class is the following:

    integral (0.5 to 1) [(x^2tan^-1(x) + 3x + 1)/ x^4 + x^2] dx

    here is a link to a picture of in case it doesn't make sense the way i typed it.
    http://img11.imageshack.us/img11/8223/mathfp6.jpg

    I really don't know how exactly to solve it. I have never done any where the inverse tangent is on the numerator. Can I get some help?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \int_{.5}^{1} \frac{x^{2}\cdot \tan^{-1} x + 3 x + 1}{x^{4} + x^{2}} \cdot dx = \int_{.5}^{1} \frac{\tan^{-1} x}{1+x^{2}}\cdot dx + \int_{.5}^{1} \frac{1 + 3 x}{x^{2}\cdot (1+x^{2})} (1)

    The integral that contains \tan^{-1} x is relatively easy to solve because is...

    \frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^{2}} (2)

    Kind regards

    \chi \sigma
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