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    I'm going through the core 3 specimin paper as I'm retaking the exam this summer and on my teachers workthrough of the question she integrates the equation in the title as 1/3 ( ln(x^3 + 2) ). What happend to the x in the x^2? I got it as x^3 /3 ( ln(x^3 + 2) ).


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    Use integration by substitution, setting u = x^3 +2. That simplifies it nicely, so the integrand is easy to evaluate, and it'll pop out exactly as your teacher has said! Hope this helps.
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    (Original post by mattjpye)
    Use integration by substitution, setting u = x^3 +2. That simplifies it nicely, so the integrand is easy to evaluate, and it'll pop out exactly as your teacher has said! Hope this helps.
    But the numerator is a derivative of the denominator, so all the OP has to do is adjust for the the 3x^2 by putting the 1/3 in.
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    If you haven't done integration by substitution, notice that the top is (sort of) the derivative of the bottom. This happens when differentiating ln(f(x)). So d(ln(x^3+2)/dx is equal to 3x^2/(x^3 +2). In your example the top isn't quite the derivative of the bottom, but if you imagine 3*1/3 x^2 on the top the 3's will cancel and you will get 1/3ln(x^3 +2)+c. To check your answer is right you can always differentiate this using the chain rule. Good luck in C3 in summer
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    Ah! Thank you all! I have done all this, I just forgot about it, this is the first time I've picked p core 3 since the exam so a few things have slipped my mind, thanks.


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