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Math Help - integration by parts

  1. #1
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    integration by parts

    I am working on a integration by parts problem.

    \int tan^{-1}(4t)dt

    Setting u=tan^{-1}(4t), then dv=dt, dv=\frac{1}{1+(4t)^2} and v=t

    Plugging into the equation, we get:

    tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt

    I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

    Can someone explain?
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  2. #2
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    Re: integration by parts

    Quote Originally Posted by baldysm View Post
    I am working on a integration by parts problem.

    \int tan^{-1}(4t)dt
    Look at this webpage.
    Thanks from topsquark
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  3. #3
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    Re: integration by parts

    Quote Originally Posted by baldysm View Post
    I am working on a integration by parts problem.

    \int tan^{-1}(4t)dt

    Setting u=tan^{-1}(4t), then dv=dt, dv=\frac{1}{1+(4t)^2} and v=t
    You have a lot of typos that make this very difficult to read. You mean
    u= tan^{-1}(4t) and dv= dt. Then du= \frac{4}{1+ (4t)^2} dt and v= t. An error that I think is NOT a typo is that you have du= \frac{1}{1+ (4t)^2} without the "4" in the numerator. You need that because of the "chain rule". To differentiate u= tan^{-1}(4t), let y= 4t so that u= tan^{-1}(y). Then \frac{du}{dy}= \frac{1}{1+ y^2} and \frac{dy}{dx}= 4. By the chain rule, \frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= \frac{1}{1+ u^2}(4)= \frac{4}{1+ 16t^2}.

    Plugging into the equation, we get:

    tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt

    I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

    Can someone explain?
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