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    I've tried solving this but to no avail.

    The first part to this question asks you to write 2/4-y^2 as partial fractions so I assume you use that for this but I just get stuck halfway through.


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    (Original post by goodwinning)
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    I've tried solving this but to no avail.

    The first part to this question asks you to write 2/4-y^2 as partial fractions so I assume you use that for this but I just get stuck halfway through.


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    Please post your working up to the point you get stuck.
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    (Original post by goodwinning)
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    Notnek got here first, don't open this.

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    Seperating variables:

    \displaystyle 

\begin{equation*}\int \frac{2}{4-y^2} \, \mathrm{d}y = \int \frac{1}{\cot x} \, \mathrm{d}x = \int \tan x \, \mathrm{d}x = \ln |\sec x| + \mathcal{C}\end{equation*}

    Then partial fractions gets us \int \frac{2 \, \mathrm{d}y}{4 - y^2} = \frac{1}{2} \int \frac{1}{y + 2} - \frac{1}{y-2} \, \mathrm{d}y

    And \int \frac{1}{y-2} \, \mathrm{d}y = \ln |y-2| and \int \frac{1}{y-2} \, \mathrm{d}y = \ln |y-2|
 
 
 
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