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    How would I integrate

    x/(1+x) dx

    I used U substitution and used u = 1+x

    So du = dx

    X = 1-u

    So I had (1-u)/u which gives

    (1/u) -1

    Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

    But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

    Hope somebody can help.

    Thanks very much,
    Ross
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    no x = u-1
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    x/(1+x) can be written as: 1 - 1/(1+x).
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    (Original post by rosschambers1987)
    How would I integrate

    x/(1+x) dx

    I used U substitution and used u = 1+x

    So du = dx

    X = 1-u

    So I had (1-u)/u which gives

    (1/u) -1

    Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

    But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

    Hope somebody can help.

    Thanks very much,
    Ross
    You don't have to use a substitution. Note that for x/(1+x), the numerator is of equal degree to the denominator, as such, you can do long division. With so few terms that should be very quick. You will then be left with some terms which you can easily integrate by inspection.
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    (Original post by rosschambers1987)
    How would I integrate

    x/(1+x) dx

    I used U substitution and used u = 1+x

    So du = dx

    X = 1-u

    So I had (1-u)/u which gives

    (1/u) -1

    Integrating that I got Ln(u) - u and substituting back in gave ln(1+x) -(1+x) which is ln(1+x) - 1 - x

    But, the answer in the AQA book and on my TI89 gives x - ln(1+x)

    Hope somebody can help.

    Thanks very much,
    Ross
    You should have put x=u-1 instead (you just rearranged u=1+x wrong), which is why your answer is -1 times what it should be. And don't worry about the 1 floating around at the end: the arbitrary constant of integration takes care of that.

    However, it's just as easy to just split \dfrac{x}{1+x} up directly. Notice that x=(1+x)-1, and so \dfrac{x}{1+x} = 1 - \dfrac{1}{1+x}. If you do this then you don't need to use a substitution.
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    use algebraic division on x/(1+x) to get 1- (1/(1+x)) and then integrate and you get the answer given
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    @nuodai: Don't you "need" a substitution to integrate \frac{1}{1+x}?
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    (Original post by DFranklin)
    @nuodai: Don't you "need" a substitution to integrate \frac{1}{1+x}?
    It should go straight to a log. I can see where you're going with what you said though... Most people doing A-Level maths have learnt the way to deal with integrals that lead to logarithms by inspection rather than substitution. Don't know if that's a bad thing or not, really.
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    Hi everyone, thanks for posting back to me. Very helpful feedback

    It's a load of questions in an AQA book where I have to choose the best method to integrate so there's a mix of all different techniques. I suppost practice on these will lead me to choose the correct method in future.

    Thanks again, most helpful!
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    (Original post by Farhan.Hanif93)
    It should go straight to a log. I can see where you're going with what you said though... Most people doing A-Level maths have learnt the way to deal with integrals that lead to logarithms by inspection rather than substitution. Don't know if that's a bad thing or not, really.
    I had no idea what people would/wouldn't know at A-level these days.
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    I only started studying A level maths in Sept last year, doing a full A level, C1, C2, C3, C4, MFP1 and MFP3.

    Done 12 and F1 and have my 3,4 and FP3 exams in June.

    Cramming! All I've learned is from the internet and some books I have.

    Managed a B in my AS and need B in the other 3 to get into Uni!
 
 
 
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