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Integration by substitution question? Watch

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    Hi, I have a query about this question on integrating using substitution.

    The question is this:
    using x = u2 , find

    \displaystyle\int \frac{1}{\sqrt x -1}\ dx


    I have got to this stage:

    \displaystyle\int \frac{2u}{u -1}\ du

    But don't know how to integrate this expression... (it doesn't appear to be a simple integration by recognition, i.e. natural log expression, and I haven't been taught how to integrate fractions otherwise)

    I'm stuck!!! Any help is much appreciated!

    Have I made a mistake?
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    Consider that 2u/(u-1) = 2(u-1)/(u-1) + 2/(u-1).
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    Have you tried dividing the numerator by the denominator.
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    (Original post by Funky_Giraffe)
    Hi, I have a query about this question on integrating using substitution.

    The question is this:
    using x = u2 , find

    \displaystyle \int \frac{1}{\sqrt x -1}\ dx


    I have got to this stage:

    \displaystyle \int \frac{2u}{u -1}\ du

    But don't know how to integrate this expression... (it doesn't appear to be a simple integration by recognition, i.e. natural log expression, and I haven't been taught how to integrate fractions otherwise)

    I'm stuck!!! Any help is much appreciated!

    Have I made a mistake?
    \displaystyle  \int \frac{2u}{u-1}\mathrm{d}u \\\\= 2\int \frac{u}{u-1}\mathrm{d}u \\\\= 2\int \frac{(u-1)+1}{u-1}\mathrm{d}u\\\\=2\int \left(\frac{u-1}{u-1}+\frac{1}{u-1}\right)\mathrm{d}u\]

    Looks better?
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    (Original post by Funky_Giraffe)


    I have got to this stage:

    \displaystyle\int \frac{2u}{u -1}\ du

    But don't know how to integrate this expression... (it doesn't appear to be a simple integration by recognition, i.e. natural log expression, and I haven't been taught how to integrate fractions otherwise)
    Use a second substitution y=u-1 \Rightarrow u=y+1
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    Split it up as much as you can. Take the numerator out after you've made your Substitutions!
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    (Original post by atsruser)
    Use a second substitution y=u-1 \Rightarrow u=y+1
    A second substitution just over complicates it (somewhat).
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    (Original post by B_9710)
    A second substitution just over complicates it (somewhat).
    You've haven't explained why you think so, but I beg to differ anyway.

    1. All A level students will be familiar with substitutions, so no new technique is required, when you approach this problem via substitution. OTOH, few A level students will be taught the specific trick suggested, and fewer still would discover it on their own.

    2. The amount of work in the two methods seems approximately identical to me, so I see no big win either way there.
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    (Original post by aymanzayedmannan)
    \[\large \int \frac{2u}{u-1}du = 2\int \frac{u}{u-1}du = 2\int \frac{(u-1)+1}{u-1}du=2\int (\frac{u-1}{u-1}+\frac{1}{u-1})du\]

    Looks better?
    Thanks all! Found this method the easiest to follow, so thank you

    Just as a side point, if I need to find out √6, by binomial expansion of (1-4x)1/2 , wouldn't the best substitution for x be x = 0.1? Because I have a question here saying using the substitution x = 0.01 for this exact same bracket. I'm working through but getting an expression in terms of √0.96. I would have thought that 0.1 would give me in terms of √0.6 and from there it's easy because 0.6 and 6 are relatable.

    Do you reckon this is a typo or is there actually a way by substituting x = 0.01?
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    (Original post by atsruser)
    You've haven't explained why you think so, but I beg to differ anyway.

    1. All A level students will be familiar with substitutions, so no new technique is required, when you approach this problem via substitution. OTOH, few A level students will be taught the specific trick suggested, and fewer still would discover it on their own.

    2. The amount of work in the two methods seems approximately identical to me, so I see no big win either way there.
    It does not make it harder I suppose but I always think it is better to integrate directly if you an rather than use a sub.
    I wasn't saying what you said was wrong, I just myself wouldn't do it. But what do I know?
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    (Original post by Funky_Giraffe)
    Thanks all! Found this method the easiest to follow, so thank you

    Just as a side point, if I need to find out √6, by binomial expansion of (1-4x)1/2 , wouldn't the best substitution for x be x = 0.1? Because I have a question here saying using the substitution x = 0.01 for this exact same bracket. I'm working through but getting an expression in terms of √0.96. I would have thought that 0.1 would give me in terms of √0.6 and from there it's easy because 0.6 and 6 are relatable.

    Do you reckon this is a typo or is there actually a way by substituting x = 0.01?
    0.6 and 6 are only relatable in this context if you multiply by sqrt(10), which is a bit of a hard thing without a calculator.

    Remember that: \display \sqrt{0.96} = \frac{\sqrt{24}}{\sqrt{25}} = \frac{2\sqrt{6}}{5}, and you can multiply by 5/2 much more easily.
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    (Original post by Funky_Giraffe)
    Thanks all! Found this method the easiest to follow, so thank you
    no worries - it was actually Zacken who taught me this trick so thank him
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    (Original post by Funky_Giraffe)
    ...
    Don't thank me. Ayman deserves all the thanks, he explained it brilliantly. :yep:
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    (Original post by B_9710)
    It does not make it harder I suppose but I always think it is better to integrate directly if you an rather than use a sub.
    I wasn't saying what you said was wrong, I just myself wouldn't do it. But what do I know?
    I agree. No need to substitute if you can split up the fraction.
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    (Original post by aymanzayedmannan)
    \[\large \int \frac{2u}{u-1}du = 2\int \frac{u}{u-1}du = 2\int \frac{(u-1)+1}{u-1}du=2\int (\frac{u-1}{u-1}+\frac{1}{u-1})du\]

    Looks better?
    Zacken, is that you? :curious:

    Edit: just seen the post saying how zacken taught you it haha
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    (Original post by DylanJ42)
    Zacken, is that you? :curious:
    No, Zacken uses \displaystyle
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    (Original post by notnek)
    No, Zacken uses \displaystyle
    Astute. Also, \mathrm{d}x instead of dx.
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    (Original post by Zacken)
    Astute. Also, \mathrm{d}x instead of dx.
    \heartsuit
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    (Original post by notnek)
    I agree. No need to substitute if you can split up the fraction.
    I can't see any significant benefit of one approach over the other, for someone who knows both.
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    (Original post by atsruser)
    I can't see any significant benefit of one approach over the other, for someone who knows both.
    There isn't really, just personal preference. I would prefer to manipulate the fraction than substitute.
 
 
 
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