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    How would you integrate

    2x^2/(1+x) ?

    Attempted to use partial fractions but not successful ?
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    (Original post by jon2016)
    How would you integrate

    2x^2/(1+x) ?

    Attempted to use partial fractions but not successful ?
    Long division. You should get \frac{2x^2 }{1+x} = 2x  - \frac{2x}{1+x} then long division on the second fraction as well.
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    substitution?

    u=1+x
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    How exactly would long division work with polynomials ? The X in the numerator cant cancel out ?

    (Original post by Zacken)
    Long division. You should get \frac{2x^2 }{1+x} = 2x  - \frac{2x}{1+x} then long division on the second fraction as well.
    The textbook does this http://imgur.com/a/ABxPl

    Not sure what method they followed ?
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    (Original post by jon2016)
    How exactly would long division work with polynomials ? The X in the numerator cant cancel out ?



    The textbook does this http://imgur.com/a/ABxPl

    Not sure what method they followed ?
    What they've done is \frac{2x^2}{1+x} = \frac{2x^2 + 2x - 2x}{1 +x} = \frac{2x(x+1) - 2x}{1+x} = \frac{2x(1+x)}{1+x} - \frac{2x}{1+x} = 2x - \frac{2x}{1+x}

    No perform the same 'trick' or 'technique' on the second fraction by 2x = 2x + 2 - 2  = 2(1+x) - 2
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    (Original post by Zacken)
    What they've done is \frac{2x^2}{1+x} = \frac{2x^2 + 2x - 2x}{1 +x} = \frac{2x(x+1) - 2x}{1+x} = \frac{2x(1+x)}{1+x} - \frac{2x}{1+x} = 2x - \frac{2x}{1+x}

    No perform the same 'trick' or 'technique' on the second fraction by 2x = 2x + 2 - 2  = 2(1+x) - 2
    wow thats amazing, I would never have thought of doing that in the exam ! thanks so much.

    Out of curiosity, are there any clear things to look for , so you know which integration technique to use ? from partial, sub and by parts ?
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    (Original post by jon2016)
    wow thats amazing, I would never have thought of doing that in the exam ! thanks so much.

    Out of curiosity, are there any clear things to look for , so you know which integration technique to use ? from partial, sub and by parts ?
    The thing is, here - long division and substitution would have worked as well as just this basic simplification. You might want to learn "long division for polynomials" so you're able to do what I did in my previous post but without it requiring any creative thought and it's a lot more systematic.

    You tend to get used to what technique to use once you've done enough practice; normally IBP is when you have a product of two functions where each is easy integrable/differentiable but put together they aren't, etc...
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    (Original post by Zacken)
    The thing is, here - long division and substitution would have worked as well as just this basic simplification. You might want to learn "long division for polynomials" so you're able to do what I did in my previous post but without it requiring any creative thought and it's a lot more systematic.

    You tend to get used to what technique to use once you've done enough practice; normally IBP is when you have a product of two functions where each is easy integrable/differentiable but put together they aren't, etc...
    What about partial fractions ?
    Could we have used partial fractions for the above question ?
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    (Original post by jon2016)
    What about partial fractions ?
    Could we have used partial fractions for the above question ?
    Well, no - you can only use partial fractions when there is more than one factor in the denominator. But all you have here is (1+x) in the denominator, which you can't do anything with, unfortunately.
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    (Original post by Zacken)
    Well, no - you can only use partial fractions when there is more than one factor in the denominator. But all you have here is (1+x) in the denominator, which you can't do anything with, unfortunately.
    what about e^ root(x) / root(x)

    My textbook does it by Integration by substitution, but the fact is has two terms, suggested to me it should be Integration by parts ?
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    (Original post by jon2016)
    what about e^ root(x) / root(x)

    My textbook does it by Integration by substitution, but the fact is has two terms, suggested to me it should be Integration by parts ?
    You can do it by parts, if you want: u = e^{\sqrt{x}} \Rightarrow \frac{\mathrn{d}u}{\mathrm{d}x} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}.

    And \mathrm{d}v = \frac{1}{\sqrt{x}} \Rightarrow v= 2\sqrt{x}.

    And that would work out fine.
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    (Original post by Zacken)
    You can do it by parts, if you want: u = e^{\sqrt{x}} \Rightarrow \frac{\mathrn{d}u}{\mathrm{d}x} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}.

    And \mathrm{d}v = \frac{1}{\sqrt{x}} \Rightarrow v= 2\sqrt{x}.

    And that would work out fine.
    ah right thanks, it didnt occur to me

    I thought about what you said about being creative

    Is this a legitimate maths rearranging technique http://i.imgur.com/i1kWglN.jpg ie multpiply numerator and denominator by 2
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    (Original post by jon2016)

    Is this a legitimate maths rearranging technique http://i.imgur.com/i1kWglN.jpg ie multpiply numerator and denominator by 2
    Very much so! You learnt about this technique in middle school, I'm sure. How do you add \frac{1}{2} + \frac{1}{4}. Well, you take the \frac{1}{2} and multiplied the numerator and denominator by 2 to get \frac{2}{4}. So that you could write \frac{1}{2} + \frac{1}{4}  = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}.

    So, to answer your question, yes: what you've done is valid.
 
 
 
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