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Help with integrating without substitution (C4) Watch

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    Hi all! I missed a week from college and doing catch up work. I have to do \int \frac{x^2}{1+x^3} without substitution.

    I was there only for lessons where we did partial fractions which can not be done with this question as far as I can see? The text book I have is not much help either, and have tried live maths too!

    I have a funny feeling it is something obvious but can not see it Thanks!
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    Does the differential of the bottom bracket help you spot something relative to the top?
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    (Original post by FattyInNeed)
    Does the differential of the bottom bracket help you spot something relative to the top?
    huh? xD Sorry my maths terminology is not great do you mean if there is something similar about the denominator and numerator? (as in the x terms)
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    Anyone else know? =/ I need to get this done today and this is only the first question in the work (they are all similar basis), and I am out all day tomorrow at an awards ceremony for young person of the year so can't really get it done as going out to celebrate after ... (I know last minute! xD)
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    (Original post by Stephhcharlene)
    Anyone else know? =/ I need to get this done today and this is only the first question in the work (they are all similar basis), and I am out all day tomorrow at an awards ceremony for young person of the year so can't really get it done as going out to celebrate after ... (I know last minute! xD)
    If you differentiate the bottom, what does that give you.

    You should also know the \frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}
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    Here's my method of doing this;

    \int \frac{x^2 }{1+x^3 } dx = \int \frac{f'(x) }{f(x) } dx = ln|f(x)|

    

f(x) = 1+x^3

f'(x)= 3x^2



= \dfrac{1}{3} \int \dfrac{3x^2 }{1+x^3 } dx

= \int \dfrac{f'(x) }{f(x) } dx

= ln \ |f(x)|  \ + c

= \dfrac{1}{3} ln|1+x^3 | +c
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    (Original post by Wednesday Bass)
    If you differentiate the bottom, what does that give you.

    You should also know the \frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}
    ... I am confused. You are doing dy/dx... which is differentiation, and I am integrating in my quesiton??

    (Original post by Brecon)
    Here's my method of doing this;

    \int \frac{x^2 }{1+x^3 } dx = \int \frac{f'(x) }{f(x) } dx = ln|f(x)|

    

f(x) = 1+x^3

f'(x)= 3x^2



= \dfrac{1}{3} \int \dfrac{3x^2 }{1+x^3 } dx

= \int \dfrac{f'(x) }{f(x) } dx

= ln \ |f(x)|  \ + c

= \dfrac{1}{3} ln|1+x^3 | +c
    Thanks!!! This helps a lot
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    (Original post by Stephhcharlene)
    ... I am confused. You are doing dy/dx... which is differentiation, and I am integrating in my quesiton??
    Integration is the inverse of differentiation.

    If you look at the solution brecon's given, you'll see he uses what I've said.
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    (Original post by Wednesday Bass)
    Integration is the inverse of differentiation.

    If you look at the solution brecon's given, you'll see he uses what I've said.
    Yeah I just noticed that as I looked more indepth! Sorry! Thankyou
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    (Original post by Brecon)
    Here's my method of doing this;
    This is just integration by substitution (u=f(x)), but without explicitly stating it.
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    Ahh cool! its all making sense to me now that i have read through it all a few times
 
 
 
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