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    Just refreshing my memory on these.

    \int \frac{3}{x} dx = 3lnx + c

    \int \frac{2}{3x} dx = \frac{2}{3}lnx + c

    \frac{d}{dx} (\frac{2}{1-x}) = \frac{2}{(1-x)^2}

    please check (I think the second one is wrong)

    edit: Also

    how would you do \int \frac{3}{5x+1} dx
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    (Original post by Prokaryotic_crap)
    Just refreshing my memory on these.







    please check (I think the second one is wrong)
    All seem correct to me.
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    All correct!
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    I swear, at some point in my C4 exam, I'm going to put my head in my hands, then scream "I don't know who I am any more"

    Thank god I only need a B.
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    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx
    Make a substitution u=5x+1.
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    (Original post by IchiCC)
    All seem correct to me.
    am I wrong or whenever you differentiate ln(ax) you always get \frac{1}{x}
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    (Original post by Gemini92)
    Make a substitution u=5x+1.
    oh yeah
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    (Original post by Prokaryotic_crap)
    am I wrong or whenever you differentiate ln(ax) you always get \frac{1}{x}
    You can rewrite ln(ax) as  lna + lnx and differentiate that. lna is a constant so it differentiates to 0 so you're left with the derivative of lnx which is 1/x.
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    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx

    3/5ln(5x+1)

    If the differential of the bottom is on the top, then its just ln of the bottom times by whatever constant you need to get it to work.
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    (Original post by Gemini92)
    Make a substitution u=5x+1.
    Actually, I don't think you'd need to do that substitution.

    you'd simply get
     5/3 ln(5x +1)

    I think...
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    (Original post by fibrebiz)
    Actually, I don't think you'd need to do that substitution.

    you'd simply get
     5/3 ln(5x +1)

    I think...
    It'd be 3/5 ln(5x+1)
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    (Original post by Gemini92)
    You can rewrite ln(ax) as  lna + lnx and differentiate that. lna is a constant so it differentiates to 0 so you're left with the derivitive of lnx which is 1/x.
    that makes sense, never thought of that. last night i was differentiating ln2x and done \frac{d}{dx}(ln2x}) = \frac{1}{2x}\times 2 = \frac{1}{x} and then I noticed I always get 1/x regardless of the coefficient, so I thought I might be doing it wrong
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    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx
    Making a substitution makes it more complicated than it needs to be. Whenever possible, try to do it by inspection.

    If you consider \dfrac{d}{dx}[\mathrm{ln}(5x+1)], we get

    \dfrac{5}{5x+1}

    This is close to what we want, but not quite, as we have a 3 in the numerator not a 5, so we can adjust the numerical factor to account for this. If we put a factor of \dfrac{3}{5} infront to get,

    \dfrac{d}{dx} [\frac{3}{5}\mathrm{ln}(5x+1)], we get

    \dfrac{3}{5x+1}, hence \int \frac{3}{5x+1} dx = \frac{3}{5}\mathrm{ln}(5x+1) +c
 
 
 
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