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    I'm stuck on question 4


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    And this is as far as I've got


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    Oh that's horrible!

    Really!

    You have separated the variables correctly but the next line is just drivel. Try substituting a couple of numbers and you will agree.

    Integrate the y term by recognising a logarithmic integral and the x term by making a substitution for the denominator or by expressing it in divided out form.
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    (Original post by SDavis123)
    And this is as far as I've got


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    Line 1 = good

    Line 2 = rubbish

    \dfrac{x}{x-1} \not= 1-x

    Ditto the y side being incorrect
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    (Original post by Mr M)
    Oh that's horrible!

    Really!

    You have separated the variables correctly but the next line is just drivel. Try substituting a couple of numbers and you will agree.

    Integrate the y term by recognising a logarithmic integral and the x term by making a substitution for the denominator or by expressing it in divided out form.
    Ah yes I see what you mean so take the 2 out of the y bit to make the top differential of the bottom so you'll get ln(y^2 +1) ?


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    To explain your misconception:

    \displaystyle \frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}
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    (Original post by SDavis123)
    Ah yes I see what you mean so take the 2 out of the y bit to make the top differential of the bottom so you'll get ln(y^2 +1) ?


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    Sort of.

    You need to deal with the 2 you just threw away.
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    Ah yh sorry, (1/2)(ln(y^2 + 1))


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    (Original post by SDavis123)
    Ah yh sorry, (1/2)(ln(y^2 + 1))


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    Yes.
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    (Original post by Mr M)
    Yes.

    I'm not sure on the x part but I said

    x/(x-1)=x(x-1)^-1

    And when I integrate that I get

    ((-x^2)/4).(x-1) that right?


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    (Original post by SDavis123)
    I'm not sure on the x part but I said

    x/(x-1)=x(x-1)^-1

    And when I integrate that I get

    ((-x^2)/4).(x-1) that right?


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    You seem to be making up your own integration rules now.

    \displaystyle \frac{x}{x-1}=\frac{x-1+1}{x-1}=1+\frac{1}{x-1} which I hope you can integrate.
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    (Original post by Mr M)
    You seem to be making up your own integration rules now.

    \displaystyle \frac{x}{x-1}=\frac{x-1+1}{x-1}=1+\frac{1}{x-1} which I hope you can integrate.
    Ermmm ok haha but how did you get

    x/(x-1)=(x-1+1)/(x-1) ?


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    (Original post by SDavis123)
    Ermmm ok haha but how did you get

    x/(x-1)=(x-1+1)/(x-1) ?


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    Because x = x - 1 + 1
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    (Original post by SDavis123)
    Ermmm ok haha but how did you get

    x/(x-1)=(x-1+1)/(x-1) ?


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    Because x-1+1 = x ???

    Edit: SNAP!
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    (Original post by davros)
    Because x-1+1 = x ???
    Snap!

    Now your edit made my snap look stupid!
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    (Original post by Mr M)
    Because x = x - 1 + 1
    Ah yh obviously can't believe I didn't see that haha


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    (Original post by Mr M)
    Snap!

    Now your edit made my snap look stupid!
    I'm trying to write maths while watching the football. I should stop now before I write something stupid!
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    (Original post by davros)
    I'm trying to write maths while watching the football. I should stop now before I write something stupid!
    Football? And there was me thinking mathematicians were clever

    And what's the next move after



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    (Original post by SDavis123)
    Football? And there was me thinking mathematicians were clever

    And what's the next move after



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    Integrate it - come on this is getting a bit painful!
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    (Original post by Mr M)
    Integrate it - come on this is getting a bit painful!
    Yh I did, did you see the picture?


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