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    Could somebody please write the method to (iv) please. I’ve done it, but I don’t know where ln comes into it. The answer is to the right of the question.
    Many thanks💗
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    (Original post by Lucybradders__)

    Could somebody please write the method to (iv) please. I’ve done it, but I don’t know where ln comes into it. The answer is to the right of the question.
    Many thanks💗
    Expand (4x-5)^2 then consider how the top can be manipulated into being the derivative of the bottom. Then use \displaystyle \int \frac{f'(x)}{f(x)}.dx = \ln |f(x)| +c rule.

    EDIT: Sorry, completely misunderstood your question and got derailed.

    What do you end up with after the substitution?
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    (Original post by Lucybradders__)
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    Could somebody please write the method to (iv) please. I’ve done it, but I don’t know where ln comes into it. The answer is to the right of the question.
    Many thanks💗
    You should have ended up needing to integrate (some multiple of) \displaystyle \int \frac{1}{u}\,du (plus some other stuff).

    Post your working.
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    (Original post by RDKGames)
    Expand (4x-5)^2 then consider how the top can be manipulated into being the derivative of the bottom. Then use \displaystyle \int \frac{f'(x)}{f(x)}.dx = \ln |f(x)| +c rule.

    EDIT: Sorry, completely misunderstood your question and got derailed.

    What do you end up with after the substitution?
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    In the step where you end up with \displaystyle \int \dfrac{u^3+5u^2}{4}\,du, you've mulitplied the numerator by u^2 when you should have divided.
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    (Original post by Lucybradders__)
    ...
    \displaystyle \frac{\frac{u+5}{4}}{u^2}=\frac{  u+5}{4}\cdot u^{-2}=\frac{u^{-2}(u+5)}{4}
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    (Original post by _gcx)
    I don't think the integral converges, but I might be being dense.
    Fair point, but not relevant to the OP. (i.e. you're clearly not supposed to worry about it for these exercises).

    OP should ignore this...
    FWIW, the log term is actually the more manageable one here. The \int 1/x \,dx = \ln |x| +C approach gives you the Cauchy Principle value (where we don't worry about what happens at x = 0, because there are +ve and -ve bits that cancel), so there's some hope of getting the right answer.

    It's the 1/u^2 term that's particularly bad; clearly \int_{-1}^1 1/u^2 \,du should either be finite and positive or infinite (because 1/u^2 is always +ve or infinite). But standard integral rules give us [-1/u]_{-1}^1 = -2...
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    (Original post by DFranklin)
    Fair point, but not relevant to the OP. (i.e. you're clearly not supposed to worry about it for these exercises).

    OP should ignore this...
    FWIW, the log term is actually the more manageable one here. The \int 1/x \,dx = \ln |x| +C approach gives you the Cauchy Principle value (where we don't worry about what happens at x = 0, because there are +ve and -ve bits that cancel), so there's some hope of getting the right answer.

    It's the 1/u^2 term that's particularly bad; clearly \int_{-1}^1 1/u^2 \,du should either be finite and positive or infinite (because 1/u^2 is always +ve or infinite). But standard integral rules give us [-1/u]_{-1}^1 = -2...
    Spoiler:
    Show

    Just seems a bit off to ignore the glaring discontinuity even in the context of an exercise. Regardless I'll delete my previous post as to not confuse the OP.
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    (Original post by DFranklin)
    Fair point, but not relevant to the OP. (i.e. you're clearly not supposed to worry about it for these exercises).
    ...except that the lower limit x = 1 maps to u = -1. That makes it difficult to evaluate any ln(u) term in the integral...
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    (Original post by old_engineer)
    ...except that the lower limit x = 1 maps to u = -1. That makes it difficult to evaluate any ln(u) term in the integral...
    Not really, you just do the "oh, it's ln|u|" fudge.
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    (Original post by DFranklin)
    Not really, you just do the "oh, it's ln|u|" fudge.
    OK fair enough
 
 
 
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