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    understanding sub u

    My book has an example that I am having trouble comprehending

    Find \int_{}{} \frac{x}{\sqrt{x+1}} dx

    "Subsititution 1: The composite function \sqrt{x+1} suggests the new variable u = x+1. You might doubt whether this choice will work becuase du=dx and the x in the numerator of the integrand is unnacounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

    \int_{}{} \frac{x}{\sqrt{x+1}} dx = \int_{}{} \frac{u-1}{\sqrt{u}} du

    ....


    How should I know to let x= u-1
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    Super Member Quacky's Avatar
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    Re: understanding sub u

    You can integrate \frac{1}{\sqrt{u}}, whereas you cannot directly integrate \frac{1}{\sqrt{x+1}}

    Integration is always a bit of guess and check. One might also consider letting x=sinh^2(u) or letting x+1=u^2 because we ultimately want to simplify the radical on the denominator to get it into a form that we can directly integrate. Will my alternative suggestions work? I have no idea (although the former would be particularly ill-advised here!), but they would remove the problem caused by the radical so they would be logical guesses. When dealing with radicals, setting u (or u^2) equal to whatever is beneath the radical is usually a wise "first choice". You never "know" they will work for certain, but that's the joy of integration!
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    Re: understanding sub u

    Quote Originally Posted by delgeezee View Post
    My book has an example that I am having trouble comprehending

    Find \int_{}{} \frac{x}{\sqrt{x+1}} dx

    "Subsititution 1: The composite function \sqrt{x+1} suggests the new variable u = x+1. You might doubt whether this choice will work becuase du=dx and the x in the numerator of the integrand is unnacounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

    \int_{}{} \frac{x}{\sqrt{x+1}} dx = \int_{}{} \frac{u-1}{\sqrt{u}} du

    ....


    How should I know to let x= u-1
    not to sound flippant, but experience is the answer to your question. Integration using the method of changing variables comes more naturally after you've done it a few times. Sometimes someone else may see a "clever" substitution that you do not. It all comes with time and practice. Note that u = x+1 allows you to divide out the given quotient so that it only requires a simple power rule antiderivative.
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    Re: understanding sub u

    Quote Originally Posted by Quacky View Post
    You can integrate \frac{1}{\sqrt{u}}, whereas you cannot directly integrate \frac{1}{\sqrt{x+1}}

    Integration is always a bit of guess and check. One might also consider letting x=sinh^2(u) or letting x+1=u^2 because we ultimately want to simplify the radical on the denominator to get it into a form that we can directly integrate. Will my alternative suggestions work? I have no idea (although the former would be particularly ill-advised here!), but they would remove the problem caused by the radical so they would be logical guesses. When dealing with radicals, setting u (or u^2) equal to whatever is beneath the radical is usually a wise "first choice". You never "know" they will work for certain, but that's the joy of integration!
    Yes, letting x+1=u^2 is demonstrated in my book as the second method for substituting with this example function. I will heed you advice, "When dealing with radicals, setting u (or u^2) equal to whatever is beneath the radical is usually a wise "first choice". "

    Quote Originally Posted by skeeter View Post
    not to sound flippant, but experience is the answer to your question. Integration using the method of changing variables comes more naturally after you've done it a few times. Sometimes someone else may see a "clever" substitution that you do not. It all comes with time and practice. Note that u = x+1 allows you to divide out the given quotient so that it only requires a simple power rule antiderivative.
    Agreed, I am will make note of this type of problem and press on.... and when I get more confortable with the substitution method I will return back to it.
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    Re: understanding sub u

    Quote Originally Posted by delgeezee View Post
    My book has an example that I am having trouble comprehending

    Find \int_{}{} \frac{x}{\sqrt{x+1}} dx

    "Substitution 1: The composite function \sqrt{x+1} suggests the new variable u = x+1. You might doubt whether this choice will work because du=dx and the x in the numerator of the integrand is unaccounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

    \int_{}{} \frac{x}{\sqrt{x+1}} dx = \int_{}{} \frac{u-1}{\sqrt{u}} du

    ....


    How should I know to let x= u-1
    To begin, as Quacky mentioned, it's simpler to have \sqrt{u}
    under the line.

    This leaves you with

    \int{\frac{xdx}{\sqrt{u}}

    and you need to substitute for xdx

    x+1=u\Rightarrow\ du=dx

    and also we change from the variable x to u by subtracting 1 from x+1

    x=u-1

    So the rewriting of the entire integral suggests itself after the initial substitution
    for the denominator.
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    Re: understanding sub u

    Good News, I retried the example problem today and after a while I was able to get the answer. As you all suggested, I found that allowing u=x+1 to be the simplest approach. I also now understand why the the numerator changed to u-1=x. It was so, to keep the fuction the same, but in terms of u.

    I am now attempting, with slight difficulty, to try the other substition example by letting  u= \sqrt{x+1}

    \int{\frac{x}{\sqrt{x+1}}dx = \int{\frac{x}{u}dx

    Now I change the Numerator in terms of u.

     x= \sqrt{x+1} * \sqrt{x+1} - 1 \Rightarrow u*u-1 \Rightarrow u^2-1


    Now I need to change the variables for the dx in terms of du but I am having troubles.

    my first inclination was that since  u = \sqrt{x+1}, then du = \frac{1}{2\sqrt{x+1}}, then replace that x within the du to \frac{1}{2\sqrt{(u^2-1)+1}} = \frac{1}{2\sqrt{u^2}} = \frac{1}{2u}

    So in the original function, \int{\frac{x}{\sqrt{x+1}}dx , the dx= 1, so I should set the dx = du???
    1 = \frac{1}{2u} , multiply both sides to isolute du in terms of u, so finally, du= 2u

    So Now I have \int{\frac{u^2-1}{u}*2udu which is what the book also has

    *********
    In regards to finding the du, can I simply say Hey, the derivative of the numerator u^2-1 is 2u and i tack it on to the back of the new integrand to get \int{\frac{u^2-1}{u}*2udu ???????
    *********


    Are these the correct logical steps I should take ?????
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    Re: understanding sub u

    Nice job so far. I think it's easier to let u^2=x+1 rather than to let u=\sqrt{x+1} - both are exactly the same thing, but the first is easier to work with. You can instantly see that u^2-1=x this way. You can also differentiate both sides easily to obtain 2u~du=dx so dx=2u~du instantly, instead of going through a complicated differentiation process.

    This leaves you with:

    \int{\frac{u^2-1}{u}}\cdot~2u~du}

    Leading to:

    \int{2u^2-2~du}
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    Re: understanding sub u

    This a painful thread to follow. (I hate the wide use of u-substitution.)
    But in this case it is useful.

    If u=x+1 then x=u-1.

    Thus \frac{x}{\sqrt{x+1}} becomes \frac{u-1}{\sqrt{u}}={u^{1/2}-u^{-1/2}

    That last bit has an easy anti-derivative.
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    Re: understanding sub u

    Quote Originally Posted by Plato View Post
    I hate the wide use of u-substitution.
    I don't see why - it's a direct method that works in a large number of situations...
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    Re: understanding sub u

    Quote Originally Posted by Plato View Post
    This a painful thread to follow. (I hate the wide use of u-substitution.)
    But in this case it is useful.

    If u=x+1 then x=u-1.

    Thus \frac{x}{\sqrt{x+1}} becomes \frac{u-1}{\sqrt{u}}={u^{1/2}-u^{-1/2}

    That last bit has an easy anti-derivative.
    I agree that u-substitution should be avoided if possible. What alternative would you suggest here?

    Quote Originally Posted by Prove It View Post
    I don't see why - it's a direct method that works in a large number of situations...
    With all due respect, it's usually one of the more time-consuming methods; I feel like people rely on it too readily when there may be numerous shorthand approaches available.
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    Re: understanding sub u

    Quote Originally Posted by Quacky View Post
    I agree that u-substitution should be avoided if possible. What alternative would you suggest here?

    With all due respect, it's usually one of the more time-consuming methods; I feel like people rely on it too readily when there may be numerous shorthand approaches available.
    I have to respectfully disagree. I feel it is better to become good at one method instead of having to try to remember numerous "shorthand" methods (in other words, formulas, like \displaystyle \begin{align*} \int{\frac{f'(x)}{f(x)}\,dx} = \ln{\left|f(x)\right|} + C \end{align*} for example). It's also better from a teaching perspective to make sure that students can understand the underlying reasoning behind these "shorthand" methods, which they can only get by becoming good at the u-substitutions.
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    Re: understanding sub u

    Quote Originally Posted by delgeezee View Post
    Good News, I retried the example problem today and after a while I was able to get the answer. As you all suggested, I found that allowing u=x+1 to be the simplest approach. I also now understand why the the numerator changed to u-1=x. It was so, to keep the fuction the same, but in terms of u.

    I am now attempting, with slight difficulty, to try the other substition example by letting  u= \sqrt{x+1}

    \int{\frac{x}{\sqrt{x+1}} \dx = \int{\frac{x}{u}dx

    Now I change the Numerator in terms of u.

     x= \sqrt{x+1} * \sqrt{x+1} - 1 \Rightarrow u*u-1 \Rightarrow u^2-1


    Now I need to change the variables for the dx in terms of du but I am having troubles.

    my first inclination was that since  u = \sqrt{x+1}, then du = \frac{1}{2\sqrt{x+1}}, then replace that x within the du to \frac{1}{2\sqrt{(u^2-1)+1}} = \frac{1}{2\sqrt{u^2}} = \frac{1}{2u}

    So in the original function, \int{\frac{x}{\sqrt{x+1}}dx , the dx= 1, so I should set the dx = du???
    1 = \frac{1}{2u} , multiply both sides to isolute du in terms of u, so finally, du= 2u

    So Now I have \int{\frac{u^2-1}{u}*2udu which is what the book also has

    *********
    In regards to finding the du, can I simply say Hey, the derivative of the numerator u^2-1 is 2u and i tack it on to the back of the new integrand to get \int{\frac{u^2-1}{u}*2udu ???????
    *********


    Are these the correct logical steps I should take ?????
    You are a little bit "heavy-handed" on the calculations.

    As mentioned by Prove It and earlier by skeeter,
    it is experience with the technique that helps you master it.

    \int{\frac{x}{\sqrt{x+1}}}\ dx

    u=\sqrt{x+1}

    We require xdx in terms of u

    u=\sqrt{x+1}\Rightarrow\ u^2=x+1\Rightarrow\ x=u^2-1

    \frac{dx}{du}=\frac{d}{du}\left(u^2-1\right)=2u

    \Rightarrow\ dx=2udu\Rightarrow\ xdx=\left(u^2-1\right)2udu

    \Rightarrow\int{\frac{x}{\sqrt{x+1}}}\ dx=\int{\frac{\left(u^2-1\right)2u}{u}}\ du=2\int{\left(u^2-1\right)}du
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