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easy integration question Watch

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    Integrate

    This is what I did:

    1. wrote it out differently
    2. added one to the power and then divided by the new power.
    3. simplified it to that

    But the mark scheme tells me it's this??
    https://gyazo.com/80314eed652540fc33d89232e1e53f98

    Please help, I'm so confused
    I'm answering question 7i in Core 3 June 2014, https://gyazo.com/82b9d54752de8a44cfe66aa1e5af18fb
    Mark scheme: https://gyazo.com/13c7a1e629a769e708f2e2ff56d384b4
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    divide by differential of (4x+1)
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    (Original post by Inker)
    Integrate

    This is what I did:

    1. wrote it out differently
    2. added one to the power and then divided by the new power.
    3. simplified it to that

    But the mark scheme tells me it's this??
    https://gyazo.com/80314eed652540fc33d89232e1e53f98

    Please help, I'm so confused
    I'm answering question 7i in Core 3 June 2014, https://gyazo.com/82b9d54752de8a44cfe66aa1e5af18fb
    Mark scheme: https://gyazo.com/13c7a1e629a769e708f2e2ff56d384b4
    But if you differentiate your answer you get 4\cdot \frac{1}{2} \cdot (2\sqrt{3}(4x+1)^{-1/2})=4\sqrt{3}(4x+1)^{-1/2} which is not what you started with. Also, you need +c.

    Anyway, you forgot to take into account that the coefficient of x comes into play.

    Just use substitution and see where you get with it.
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    Since the bracket is (ax+1), when integrating, you have to multiply your integral by 1/a. Hence the answer being the same as yours but divided by 4.
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    Omg, thank you so so much. I need to be way more careful from now on.
    (wanted to rep RDKGames but it said I've repped you too recently)
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    Have to use chain rule in reverse, so divide by 1/2 but also divide by the differential of the brackets.
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    (4x+1)^-1/2= (4x+1)^1/2 when integrated

    Differentiate the new (4x+1)1/2
    to get 1/2(4x+1)^1/2

    Times tat by root 3 and you get your answer

    Look up examsolutions reverse chain rule for help

    Look at this for example

    https://www.youtube.com/watch?v=QstcvzypmuM
 
 
 
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