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    • Thread Starter
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    Not sure about this question:
    \int\frac{3-5x}{(1-x)(2-3x)}dx

    Books has answer as:
    ln|\frac{2-3x}{(1-x)^2}|+c

    However I only got to, and couldn't get to the given answer from:
    -2ln|1-x|+\frac{ln|2-3x|}{3}+c

    Not sure if book is wrong or not so could somebody check please, thanks
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    Your answer is correct.
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    Yeah you got the right answer. Well done fellow mathematician. Enjoy that feeling you get when you get an answer where the book says it's wrong but you find out you are actually right.
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    (Original post by Mr M)
    Your answer is correct.
    Likewise (just a sign here though):

    Question:
    \int\frac{17-5x}{(3+2x)(2-x)^2}dx

    Book:
    ln|\frac{3+2x}{2-x}|+\frac{1}{2-x}+c

    My answer:
    ln|\frac{3+2x}{2-x}|-\frac{1}{2-x}+c

    Cheers.
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    (Original post by fayled)
    Likewise (just a sign here though):

    Question:
    \int\frac{17-5x}{(3+2x)(2-x)^2}dx

    Book:
    ln|\frac{3+2x}{2-x}|+\frac{1}{2-x}+c

    My answer:
    ln|\frac{3+2x}{2-x}|-\frac{1}{2-x}+c

    Cheers.
    Your answer is right.
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    Yet another (hope you don't mind me quoting you - saves making a new thread about the same thing) but this is by substitution this time. Book seems to have gained a u term somewhere...

    Question:
    \int\frac{\sqrt{x^2+4}}{x}dx using the substitution u^2=x^2+4

    Book:
    \sqrt{x^2+4}+ln|\frac{\sqrt{x^2+  4}-2}{\sqrt{x^2+4}+2}|+c

    My answer:
    ln|\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}|+c

    Thanks.
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    (Original post by fayled)
    ..
    You haven't quoted me and you've made a mistake typing that out. It is completely unnecessary to make a substitution here anyway (unless the question is wrong too).

    Edit: Question has changed - give me a minute.
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    (Original post by fayled)
    ...
    Your answer is wrong. Do you want to show some working?

    Edit: I suspect you have messed up the partial fractions. Did you notice the improper fraction?
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    (Original post by Mr M)
    Your answer is wrong. Do you want to show some working?

    Edit: I suspect you have messed up the partial fractions. Did you notice the improper fraction?
    Ugh sorry about that, made a hash of it at first.

    And that's exactly where I went wrong, thanks.
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    (Original post by Mr M)
    Your answer is wrong. Do you want to show some working?

    Edit: I suspect you have messed up the partial fractions. Did you notice the improper fraction?
    If you integrate by substitution with limits and the substitution is for example u2=2x-1, then when I substitute in the two x limits, I will get two corresponding u values for each x limit due to the square root (i.e if x=1 then u=+/-1), which gives four possible combinations of limits in u, so which pairs would I use? This has confused me. Instinctively I would just pick the positive ones but I feel like I am being stupid.
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    (Original post by fayled)
    If you integrate by substitution with limits and the substitution is for example u2=2x-1, then when I substitute in the two x limits, I will get two corresponding u values for each due to the square root, which gives four possible combinations of limits in u, so which pairs would I use? This has confused me. Instinctively I would just pick the positive ones but I feel like I am being stupid.
    Positive is the way to go.
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    (Original post by Mr M)
    Positive is the way to go.
    Right thanks. Why is it that it works that way though?

    I suppose it would be clearer if I found the integral, put it back in terms of x and then used the limits wouldn't it?
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    (Original post by fayled)
    Right thanks. Why is it that it works that way though?

    I suppose it would be clearer if I found the integral, put it back in terms of x and then used the limits wouldn't it?
    You need to think about what you're trying to do! Your aim should be to write u = f(x) where f is a "proper" single-valued function, and this dictates what the correct limits for u will be.

    In this case you have a choice between u = \sqrt{2x-1} and u = -\sqrt{2x-1}. If you choose the second, you get negative limits and an extra minus sign for du.
 
 
 
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