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     \frac{ \sqrt(x^2+4)}{x} dx

    use the substitution  u^2 = x^2 + 4
    ?
    I'm confused. Does the integral have to be converted completely in terms of u? What does it go to before actual integration?
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    (Original post by lollage123)
     \frac{ \sqrt(x^2+4)}{x} dx

    use the substitution  u^2 = x^2 + 4
    ?
    I'm confused. Does the integral have to be converted completely in terms of u? What does it go to before actual integration?
    Yes. You need to differentiate what you will use for your substitution (implicitly in this case) and find dx in terms of du.

    Once you've done that, make the substitution and you'll find yourself with an integral completely in terms of u.

    See what you can do
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    (Original post by Indeterminate)
    Yes. You need to differentiate what you will use for your substitution (implicitly in this case) and find dx in terms of du.

    Once you've done that, make the substitution and you'll find yourself with an integral completely in terms of u.

    See what you can do

    udu/x = dx

    I thought the dx part had to be entirely in terms of u as well? I have

     \frac {u}{u^2 - 4} \times \frac{udu}{x}

    Its not in terms of u completely so what do i do?
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    (Original post by lollage123)
    udu/x = dx

    I thought the dx part had to be entirely in terms of u as well? I have

     \frac {u}{u^2 - 4} \times \frac{udu}{x}

    Its not in terms of u completely so what do i do?
    No.

    \dfrac{du}{dx} = \dfrac{x}{u}

    Get du in terms of dx, and write

    \dfrac{1}{x} = \dfrac{x}{x^2}

    Then substitute everything in.
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    (Original post by Indeterminate)
    No.

    \dfrac{du}{dx} = \dfrac{x}{u}

    Get du in terms of dx, and write

    \dfrac{1}{x} = \dfrac{x}{x^2}

    Then substitute everything in.
    I still don't understand, I'm getting

     dx = \frac {udu}{x}

    [tex] du = \frac {xdx}{du} [/tex}

    How would using du help? and why sub into that expression?
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    (Original post by lollage123)
    I still don't understand, I'm getting

     dx = \frac {udu}{x}

    [tex] du = \frac {xdx}{du} [/tex}

    How would using du help? and why sub into that expression?
    everything MUST end up in terms of u.

    You have a u on top from the square root and this leaves you with dx/x which is the same as udu/(x^2). Write x^2 in terms of u and you now have the whole integral in terms of u.
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    (Original post by lollage123)
    I still don't understand, I'm getting

     dx = \frac {udu}{x}

     du = \frac {xdx}{du}

    How would using du help? and why sub into that expression?
    So this is a fine mixture of u and x variables and du and dx integrating factors.
    Using \displaystyle u^2=x^2+4 substitution
    for you integral will be
    \displaystyle \sqrt{x^2+4}=u
    \displaystyle x=\sqrt{u^2-4}
    and from this
    \displaystyle \frac{dx}{du}=\frac{u}{\sqrt{u^2-4}}
    so
    \displaystyle dx=\frac{u}{\sqrt{u^2-4}}\cdor du
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    (Original post by ztibor)
    So this is a fine mixture of u and x variables and du and dx integrating factors.
    Using \displaystyle u^2=x^2+4 substitution
    for you integral will be
    \displaystyle \sqrt{x^2+4}=u
    \displaystyle x=\sqrt{u^2-4}
    and from this
    \displaystyle \frac{dx}{du}=\frac{u}{\sqrt{u^2-4}}
    so
    \displaystyle dx=\frac{u}{\sqrt{u^2-4}}\cdor du
    Sorry I don't get it.

    Here's my working:

     u^2 =x^2 + 4
    differentiating implicitly:
     2u \frac {du}{dx} = 2x
     u\frac{du}{dx} =  x
     udu = x dx
     x = \sqrt(u^2 -4)
     u = \sqrt (x^2 + 4)
    Integral becomes
     \frac {u}\sqrt (u^2 - 4)
    But how do I put the converted form of dx in? as I can't get an x on top to go with the dx, because there is u on top
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    Okay I understand getting the integral to

     \frac {u}{x} \times \frac {udu}{x}

    But what I didn't know is that you could convert the integral into u and x terms to get an x term (x^2) that you know in u. How would you spot this?

    Thanks all
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    (Original post by lollage123)
     \frac{ \sqrt(x^2+4)}{x} dx
    I'm confused. Does the integral have to be converted completely in terms of u? What does it go to before actual integration?
    Yes.

    when you have udu=xdx go back to your original substitution and get x in terms of u.

    therefore dx= u/x du and since x = (u^(2)-4)^(1/2) dx=u/((u^(2)-4)^(1/2)) then substitute.

    then you should be able to solve it
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    (Original post by lollage123)
    Okay I understand getting the integral to

     \frac {u}{x} \times \frac {udu}{x}

    But what I didn't know is that you could convert the integral into u and x terms to get an x term (x^2) that you know in u. How would you spot this?

    Thanks all
    Sorry if I am being thick but with the x^2 at the bottom can you not just sub in u^2 - 4? I am finding this difficult as well
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    (Original post by lollage123)
    Okay I understand getting the integral to

     \frac {u}{x} \times \frac {udu}{x}

    But what I didn't know is that you could convert the integral into u and x terms to get an x term (x^2) that you know in u. How would you spot this?

    Thanks all
    Multiply the fraction together, then express the x2 in terms of u.
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    (Original post by Goods)
    Yes.

    when you have udu=xdx go back to your original substitution and get x in terms of u.

    therefore dx= u/x du and since x = (u^(2)-4)^(1/2) dx=u/((u^(2)-4)^(1/2)) then substitute.

    then you should be able to solve it
    yes but then the integration using the substitution is even more complicated than the original and using the substitution is pretty much pointless no?
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    (Original post by kvohra)
    yes but then the integration using the substitution is even more complicated than the original and using the substitution is pretty much pointless no?
    yeah your right. i wasn't thinking about the question as a whole. instead it would be better to have dx =u^(2)/u^(2)-4 du or with partial fractions 1 - 4/(u^(2) -4) which is easier to solve.
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    (Original post by Goods)
    yeah your right. i wasn't thinking about the question as a whole. instead it would be better to have dx =u^(2)/u^(2)-4 du or with partial fractions 1 - 4/(u^(2) -4) which is easier to solve.
    (Original post by kvohra)
    yes but then the integration using the substitution is even more complicated than the original and using the substitution is pretty much pointless no?
    Both methods come up with the integral that Goods has stated, resolution into partial fractions then provides the correct integral.
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    (Original post by Goods)
    yeah your right. i wasn't thinking about the question as a whole. instead it would be better to have dx =u^(2)/u^(2)-4 du or with partial fractions 1 - 4/(u^(2) -4) which is easier to solve.
    Oh, now I get it, cheers!
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    (Original post by joostan)
    Both methods come up with the integral that Goods has stated, resolution into partial fractions then provides the correct integral.
    I just got stuck but now I see it
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    By the way for integrating by substitution, how do you know what substitution to use? Is there a general rule?

    What would you use for

    x(2+x)^3 and why? I chose (2+x)^4 but I don't really know why and it sort of got complicated..
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    (Original post by lollage123)
    By the way for integrating by substitution, how do you know what substitution to use? Is there a general rule?

    What would you use for

    x(2+x)^3 and why? I chose (2+x)^4 but I don't really know why and it sort of got complicated..
    Let u = 2+x

    Now ur equation becomes (u-2)(u)^3 du. U can easily multiply that in and integrate.

    A general rule for an equation like that is that u substitute the one within the bracket that has the indicy. E.g (5x+4)^5 let u = 5x+4. Do not include the indicy though
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    (Original post by lollage123)
    By the way for integrating by substitution, how do you know what substitution to use? Is there a general rule?

    What would you use for

    x(2+x)^3 and why? I chose (2+x)^4 but I don't really know why and it sort of got complicated..
    Besides integration by parts and expanding using Pascal's triangle, you could choose

    u=2+x

    However, they won't ask you to devise a suitable substitution in C4.
 
 
 
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