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    Hi,

    So I've been trying to get my head around integration in c3 before i dive into it next year and came across this question

    "The curve y=f(x) passes through the point (1,-3)
    given that f'(x) = (2x-1)^2 divided by x, find an expression for f(x)"

    The thing that stumps me about me is how there is x's on the top and bottom of the fraction. i was thinking of approaching it using f'(x)/f(x)dx=ln|f(x)| +c

    but then i dont know what common factor can be taken out so that if the demonimater is to be differentiated it would be the numerator.

    another method i thought of is to separate the fraction to (2x-1)^2 *1/x so then i can get ln|x| and (2x-1)^3/6 +c

    But according to Mathway it seems thats not correct so can someone point out what mistake I've done to make myself puzzled? Thanks
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    (Original post by Fruitbasket786)
    Hi,

    So I've been trying to get my head around integration in c3 before i dive into it next year and came across this question

    "The curve y=f(x) passes through the point (1,-3)
    given that f'(x) = (2x-1)^2 divided by x, find an expression for f(x)"

    The thing that stumps me about me is how there is x's on the top and bottom of the fraction. i was thinking of approaching it using f'(x)/f(x)dx=ln|f(x)| +c

    but then i dont know what common factor can be taken out so that if the demonimater is to be differentiated it would be the numerator.

    another method i thought of is to separate the fraction to (2x-1)^2 *1/x so then i can get ln|x| and (2x-1)^3/6 +c

    But according to Mathway it seems thats not correct so can someone point out what mistake I've done to make myself puzzled? Thanks
    It is not of the form f'(x)/f(x). Something like that would be eg integrating \frac{cosx}{sinx} or more similar to your example, \frac{2x+1}{x^2 + x - 5}.

    I'm not sure what you mean by separating - at least, I see what you're trying to do but you can't integrate two things a and b when it's ab, to give the integral of a * the integral of b. This is not necessarily true.

    One thing you can do is to look at \frac{(2x-1)^2}{x} and think about a technique you'll most likely have used at AS to manipulate things.
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    (Original post by Fruitbasket786)
    Hi,

    So I've been trying to get my head around integration in c3 before i dive into it next year and came across this question

    "The curve y=f(x) passes through the point (1,-3)
    given that f'(x) = (2x-1)^2 divided by x, find an expression for f(x)"

    The thing that stumps me about me is how there is x's on the top and bottom of the fraction. i was thinking of approaching it using f'(x)/f(x)dx=ln|f(x)| +c

    but then i dont know what common factor can be taken out so that if the demonimater is to be differentiated it would be the numerator.

    another method i thought of is to separate the fraction to (2x-1)^2 *1/x so then i can get ln|x| and (2x-1)^3/6 +c

    But according to Mathway it seems thats not correct so can someone point out what mistake I've done to make myself puzzled? Thanks
    Or you could just expand he bracket on the numerator and separate that one whole fraction into 3 different ones with x as the denominator for each. (or just go ahead and divide top and bottom by x; giving you a new denominator of 1 so you can ignore it.) 2 of them will divide and integrate nicely, and the other you will need to integrate 1/x but you should know that goes into ln(x). Put a +c at the end and job done. Don't overthink this.

    The \frac{f'(x)}{f(x)} rule doesn't apply here because x does not differentiate to give (2x-1)^2
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    (Original post by RDKGames)
    Or you could just expand he bracket on the numerator and separate that one whole fraction into 3 different ones with x as the denominator for each. 2 of them will divide and integrate nicely, and the other you will need to integrate 1/x but you should know that goes into ln(x). Put a +c at the end and job done. Don't overthink this.

    The \frac{f'(x)}{f(x)} rule doesn't apply here because x does not differentiate to give (2x-1)^2
    Didnt realise it would be that straight forward, thanks for pointing that out for me! I would've been sitting for ages trying to do some absurd method to solve it xD
 
 
 
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