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    how do i integrate y=1296/(y+3)^4 ??
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    (Original post by elvss567)
    how do i integrate y=1296/(y+3)^4 ??
    Do you mean \displaystyle y=\frac{1296}{(x+3)^4} ??

    Just use a substitution.
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    (Original post by RDKGames)
    Do you mean \displaystyle y=\frac{1296}{(x+3)^4} ??

    Just use a substitution.
    yes and substitution as in lnx or substitution as in 1296(x+3)^-4??
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    (Original post by elvss567)
    yes and substitution as in lnx or substitution as in 1296(x+3)^-4??
    Substitution as in u=x+3
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    (Original post by RDKGames)
    Substitution as in u=x+3...
    could you possibly provide step by step working out?
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    If you mean:

     \displaystyle \int \frac{1296}{(x+3)^{4}} \, dx,

    then re-write you equation where (as an example) :

     \displaystyle \frac{1}{x} = x^{-1},

    and note:

     \displaystyle \int (ax+b)^{m} \, dx \, = \, \left( \frac{1}{a(m+1)} \right) (ax+b)^{(m+1)} + C.

    Can you take it from here?
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    (Original post by elvss567)
    could you possibly provide step by step working out?
    No, but surely you've come across integration by substitution?

    \displaystyle \int 1296(x+3)^{-4}.dx = \int 1296u^{-4}.du when u=x+3
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    (Original post by simon0)
    If you mean:

     \displaystyle \int \frac{1296}{(x+3)^{4}} \, dx,

    then re-write you equation as :

     \displaystyle \frac{1}{x} = x^{-1},

    and note:

     \displaystyle \int (ax+b)^{m} \, dx \, = \, \frac{1}{a(m+1)} (ax+b)^{(m+1)} + C.

    Can you take it from here?
    yes that's great! Thankyou!!!
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    RDKGames's tip is good as well.

    What is the answer did you end up with?
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    (Original post by simon0)
    RDKGames's tip is good as well.

    What is the answer did you end up with?
    i used a mixture of both of your suggestions to get:
    -5184(x+3)^-3 /-3 + c
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    (Original post by elvss567)
    i used a mixture of both of your suggestions to get:
    -5184(x+3)^-3 /-3 + c
    How did you get 5184?
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    (Original post by RDKGames)
    How did you get 5184?
    i multiplied 1296 by -4?
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    (Original post by elvss567)
    i multiplied 1296 by -4?
    What's the reason behind this?
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    (Original post by RDKGames)
    What's the reason behind this?
    i assume when you integrate normally you have to multiply by the power and then divide by the power. e.g: 2u^2 will be 4u^3?
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    (Original post by elvss567)
    i assume when you integrate normally you have to multiply by the power and then divide by the power. e.g: 2u^2 will be 4u^3?
    The assumption is incorrect. From the very basics of C1 integration you should know that you raise the power and THEN divide by the new power on simple expressions like these.

    E.g. \displaystyle \int 2u^2 .du = \frac{2u^3}{3} +c

    Your example is flawed as since integration is opposite operation to differentiation, differentiating 4u^3 needs to give 2u^2. But \frac{d}{du}(4u^3)=12u^2 \neq 2u^2
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    (Original post by RDKGames)
    The assumption is incorrect. From the very basics of C1 integration you should know that you raise the power and THEN divide by the new power on simple expressions like these.

    E.g. \displaystyle \int 2u^2 .du = \frac{2u^3}{3} +c

    Your example is flawed as since integration is opposite operation to differentiation, differentiating 4u^3 needs to give 2u^2. But \frac{d}{du}(4u^3)=12u^2 \neq 2u^2
    Great explanation! so basically it should be :
    -3888(y+3)^-3 / -3 + c instead?
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    (Original post by elvss567)
    Great explanation! so basically it should be :
    -3888(y+3)^-3 / -3 + c instead?
    No. It should be -432(x+3)^{-3}+c. Again, no idea where you pulled your -3888 from!
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    (Original post by elvss567)
    i assume when you integrate normally you have to multiply by the power and then divide by the power. e.g: 2u^2 will be 4u^3?
    - That is not how you integrate such expressions,

    - Did you use the rule I gave you?
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    (Original post by RDKGames)
    No. It should be -432(x+3)^{-3}+c. Again, no idea where you pulled your -3888 from!
    I think i got mixed up with differentiating and integrating rules. That's the answer the mark scheme states as well.
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    (Original post by simon0)
    - That is not how you integrate such expressions,

    - Did you use the rule I gave you?
    yes,i've been multiplying by the power instead of just adding one to the power and dividing by the power, that's probably why i ended with the incorrect answers
 
 
 
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