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Math Help - Stuck on Algebra in an Integral

  1. #1
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    Stuck on Algebra in an Integral

    I know that this is an integral problem, but I'm stuck on some algebra within it. I'm doing a u-substitution on this problem:

    \int \frac{\sqrt{x}}{x^2+x} dx

    \displaystyle let u = \sqrt{x} \rightarrow du = \frac{dx}{2\sqrt{x}} \rightarrow 2 du = x^\frac{-1}{2} dx

    So far, so good. Here's where I am stuck. According to Wolfram Alpha, when I do my u-substitution, I should get:

    2 \int \frac{u^2}{u^4 + u^2} du

    The denominator and the constant are obvious, but I can't see how they get from 2 du = x^\frac{-1}{2} dx to u^2 du in the numerator. Can somebody show this to me? Clearly I have a hole in my algebra abilities.

    Thanks.
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  2. #2
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    Re: Stuck on Algebra in an Integral

    du = dx/2x^1/2 = dx/2u
    So dx = 2u*du
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  3. #3
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    Re: Stuck on Algebra in an Integral

    Quote Originally Posted by joatmon View Post
    I know that this is an integral problem, but I'm stuck on some algebra within it. I'm doing a u-substitution on this problem:

    \int \frac{\sqrt{x}}{x^2+x} dx

    \displaystyle let u = \sqrt{x} \rightarrow du = \frac{dx}{2\sqrt{x}} \rightarrow 2 du = x^\frac{-1}{2} dx

    So far, so good. Here's where I am stuck. According to Wolfram Alpha, when I do my u-substitution, I should get:

    2 \int \frac{u^2}{u^4 + u^2} du

    The denominator and the constant are obvious, but I can't see how they get from 2 du = x^\frac{-1}{2} dx to u^2 du in the numerator. Can somebody show this to me? Clearly I have a hole in my algebra abilities.

    Thanks.
    What I would do first is multiply top and bottom by \displaystyle 2\sqrt{x}, to give

    \displaystyle \int{\frac{\sqrt{x}}{x^2 + x}\,dx} = 2\int{ \left[  \frac{ \left(  \sqrt{x} \right)^2}{\left(\sqrt{x}\right)^4 + \left(\sqrt{x}\right)^2} \right] \left(\frac{dx}{2\sqrt{x}} \right)}

    Then make the substitution \displaystyle u = \sqrt{x} \implies du = \frac{dx}{2\sqrt{x}} and the integral becomes

    \displaystyle \begin{align*} 2\int{ \left[  \frac{ \left(  \sqrt{x} \right)^2}{\left(\sqrt{x}\right)^4 + \left(\sqrt{x}\right)^2} \right] \left(\frac{dx}{2\sqrt{x}} \right)} &= 2\int{\frac{u^2}{u^4 + u^2}\,du} \\ &= 2\int{\frac{1}{u^2 + 1}\,du} \\ &= 2\arctan{(u)} + C \\ &= 2\arctan{\left(\sqrt{x}\right)} +  C\end{align*}
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    Re: Stuck on Algebra in an Integral

    Right. That's what I got. So, how can Wolfram get u^2 in the numerator. Using u = sqrt(x), their u-substitution comes back as:

    \int \frac{u^2}{u^4 + u^2} dx

    So, a sqrt(x) comes back as u^2 in the numerator, but an x in the denominator also comes back as u^2????? This makes absolutely no sense to me. I've used Wolfram enough to trust what they say, but this one has me totally confused.
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    Re: Stuck on Algebra in an Integral

    Quote Originally Posted by joatmon View Post
    Right. That's what I got. So, how can Wolfram get u^2 in the numerator. Using u = sqrt(x), their u-substitution comes back as:

    \int \frac{u^2}{u^4 + u^2} dx

    So, a sqrt(x) comes back as u^2 in the numerator, but an x in the denominator also comes back as u^2????? This makes absolutely no sense to me. I've used Wolfram enough to trust what they say, but this one has me totally confused.
    Did you read my post? The numerator is \displaystyle x = \left(\sqrt{x}\right)^2 = u^2 since \displaystyle u = \sqrt{x}...
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    Re: Stuck on Algebra in an Integral

    Yes, I did read it. Thanks for posting it. Right now, however, we are working on partial fractions, and I see a hole in my ability to solve problems of this type, so I am trying to address that specific issue. Your approach is very clever, and in another context, I would happily use it. I need to focus on this issue in the event that I get something of this nature on a test.
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    Re: Stuck on Algebra in an Integral

    Quote Originally Posted by joatmon View Post
    Yes, I did read it. Thanks for posting it. Right now, however, we are working on partial fractions, and I see a hole in my ability to solve problems of this type, so I am trying to address that specific issue. Your approach is very clever, and in another context, I would happily use it. I need to focus on this issue in the event that I get something of this nature on a test.
    The point that I'm making is that this is the approach that Wolfram uses...
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