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    How would you approach this question??

    I looked at the mark scheme and don't understand what it has done. It uses natural logs.

    maybe a pointer in the correct direction to what to look up but also explanation would help.

    Thanks!
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    (Original post by Xphoenix)
    How would you approach this question??

    I looked at the mark scheme and don't understand what it has done. It uses natural logs.

    maybe a pointer in the correct direction to what to look up but also explanation would help.

    Thanks!
    Sub u = denominator
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    (Original post by RDKGames)
    Factor out the 9 and you have x^2 + 2 on the numerator. You can manipulate this into 3(x^2+2) while taking a third out of the whole integral. The result inside the integral will be the numerator being the differential of the denominator thus natural logs.

    I don't get this part, like Why did you do this? (See bold)
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    (Original post by Xphoenix)
    I don't get this part, like Why did you do this? (See bold)
    For the integral to hold, whatever you do to the to the numerator then you must do the exact opposite to it as well. Since you are multiplying by 3, you must also divide it; hence the third. 3 and a third are constants, therefore you can move whichever (or both) inside and outside the integral freely. You may think this is pointless, but multiplication by 3 gets it into the wanted form of f'(x) over f(x) which is easy to integrate.
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    (Original post by Xphoenix)
    I don't get this part, like Why did you do this? (See bold)
    let f(x) = x^3 + 6x
    f'(x) = 3x^2 + 6
    3*f'(x) = the numerator.
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    Just note that 9x^2 + 18 = 3(3x^2 + 6) so your integral is \displaystyle 3\int \frac{3x^2 + 6}{x^3 + 6x} \, \mathrm{d}x which is a standard form.
 
 
 
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