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    Dunno how to integrate this, I've tried everything I know to date about using substitution but I have this feeling there's a different technique I haven't met yet as I'm self teaching.

     \int \frac {\sqrt{x^2 + 4}}x   dx

     u^2 = x^2 + 4 is the substitution given.

    Any help appreciated, thanks.
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    well, what is your working?
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    something you missed is that you are going to integrate with respect to x i.e. dx.

    Therefore you also need to differentiate u^2 = x^2 + 4 to get get an equation in the form:

    du/dx = ....

    then make dx is the subject and substitute that equation into your integral.
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    If you've been given that substitution, obviously you can replace the numerator with u

    The next thing to do is rearrange the expression u^2 = x^2 + 4 and find an expression for x, and replace the denominator with it.

    Now, differentiate the expression u^2 = x^2 + 4 with respect to x, and rearrange it to find an expression for dx in terms of u, and replace the dx at theend of the integral with this expression.


    And now you should have something which is easy to integrate. Looks like it might be an inverse trig function or something.
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    (Original post by Totally Tom)
    well, what is your working?
    This is my working,

    I replaced the numerator with u , because u^2 = x^2 + 4 and the denominator with \sqrt {u^2 - 4} as found an expression in terms of u. I found du/dx and made du the subject and this is what I get u^2 /(\sqrt {u^2 - 4})^2 du .

    I have a feeling all that is wrong.
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    (Original post by gman2k8)
    This is my working,

    I replaced the numerator with u , because u^2 = x^2 + 4 and the denominator with \sqrt {u^2 - 4} as found an expression in terms of u. I found du/dx and made du the subject and this is what I get u^2 /(\sqrt {u^2 - 4})^2 du .

    I have a feeling all that is wrong.
    i agree with this.

    sort out the fraction and integrate.
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    could you write the bottom of the fraction as (u+2)(u-2) and write the top as (u+2)(u-2)+4 so you will have 1+4/(u^2-4) making it easier to integrate?
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    (Original post by j.d.)
    could you write the bottom of the fraction as (u+2)(u-2) and write the top as (u+2)(u-2)+4 so you will have 1+4/(u^2-4) making it easier to integrate?
    there's no need to factorise it first.
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    Now use Partial Fraction, I think .
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    no need for partials ... use another sub let y be (u^2-4) then integrate for a log
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    lol use partials
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    What?

    Spoiler:
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    Express \dfrac{u^2}{u^2 - 4} as 1 + \dfrac{4}{u^2 - 4}
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    yeah. then the fraction bit in partials is 1/u-2 - 1/u+2
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    Thanks for the help guys, I see why I wasn't getting anywhere, I havent done partial fractions or anything like that yet.
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    (Original post by j.d.)
    yeah. then the fraction bit in partials is 1/u-2 - 1/u+2
    You really should use brackets.

    1/(u-2) - 1/(u+2)
 
 
 
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