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    If  \displaystyle \int \frac{1}{4-x^2} dx = \frac{1}{2} artanh \frac{x}{2} +c =\frac{1}{4} \ln \frac{x+2}{2-x}}

    Why doesn't  \displaystyle \int \frac{-1}{4-x^2} dx =-\frac{1}{2} artanh \frac{x}{2} + c =\frac{1}{4} \ln \frac{2-x}{x+2} +c ?
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    Anyone?????????
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    (Original post by Ano123)
    Anyone?????????
    how do you know that it is wrong ?

    :beard:
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    Cuz you can't distribute negative like that (3rd step)
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    (Original post by GUMI)
    Cuz you can't distribute negative like that (3rd step)
    Why not?
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    (Original post by the bear)
    how do you know that it is wrong ?

    :beard:
    It doesn't give the same result if you do it by partial fractions (it does work without the negative - it works fine for both methods)
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    (Original post by Ano123)
    If  \displaystyle \int \frac{1}{4-x^2} dx = \frac{1}{2} artanh \frac{x}{2} +c =\frac{1}{4} \ln \frac{x+2}{2-x}}

    Why doesn't  \displaystyle \int \frac{-1}{4-x^2} dx =-\frac{1}{2} artanh \frac{x}{2} + c =\frac{1}{4} \ln \frac{2-x}{x+2} +c ?
    I think maybe because  |2-x| = |x-2| (giving you what is in the formula booklet) so:

     | \dfrac{2-x}{x+2} | = | \dfrac{x-2}{x+2} |

    So they both work (I think).
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    (Original post by Ano123)
    Anyone?????????
    Who say's it's incorrect? That's perfectly fine. Just remember your modulus signs.
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    (Original post by Zacken)
    Who say's it's incorrect? That's perfectly fine. Just remember your modulus signs.
    Yeah I figured that it had to be correct haha.
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    It definitely works. You can simply factor out -1.
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    When I do  \int \frac{-6}{9-x^2} dx using hyperbolics i get   \ln \frac{u-3}{u+3}
    When I do it by partial fractions I get   \ln \frac{3-u}{u+3}
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    Is  \displaystyle \int \frac{1}{1-x^2} dx =\frac{1}{2} \ln \left \mid \frac{1+x}{1-x} \right \mid .
    Is there an absolute inside the log?
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    (Original post by Ano123)
    When I do  \int \frac{-6}{9-x^2} dx using hyperbolics i get   \ln \frac{u-3}{u+3}
    When I do it by partial fractions I get   \ln \frac{3-u}{u+3}
    You don't get either of those, slap on those modulus signs!
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    (Original post by Ano123)
    Is  \displaystyle \int \frac{1}{1-x^2} dx =\frac{1}{2} \ln \left \mid \frac{1+x}{1-x} \right \mid .
    Is there an absolute inside the log?
    Yes, you can't have the logarithm of a negative number in the real plane.
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    (Original post by Ano123)
    When I do  \int \frac{-6}{9-x^2} dx using hyperbolics i get   \ln \frac{u-3}{u+3}
    When I do it by partial fractions I get   \ln \frac{3-u}{u+3}
    same answer
    you need mods
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    Damn. Makes sense now.
    I always used to put the mod signs but for some reason I started thinking that it wasn't right like that.
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    (Original post by TeeEm)
    same answer
    you need mods
    How you doing sir
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    (Original post by It'sDuckFam)
    How you doing sir
    very good ..
    I am teaching now
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    Is this further maths
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    (Original post by under8ed)
    Is this further maths
    Yes.
    Well, hyperbolic functions only appear at further, partial fractions is core 4
 
 
 
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