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    Firstly how could (x^2+1)/(x(x-1)) be re-written as 1+ (x+1)/(x(x-1))

    and then how could (x+1)/(x(x-1)) be written as partial fractions. I'm guessing it would be something like A/(x)+ B/(x-1). Thanks for your help
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    For your first question, if you add and take away an x from the top of the equation (obviously having no overall effect) then you should be able to see that you can factorise a bit to obtain an x(x-1), which can be separated and cancelled to the 1+ that you require.

    Secondly, you're right with the idea for A and B. Do this, multiply both sides by the bottom of your fraction and compare coefficients to get values for A and B

    Hope this helps
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    For the first case: since the highest degree of x on the top is 2 and the highest degree of x on the bottom is also 2, its a good idea to long divide it first.
    Thus :

    \dfrac {x^2+1}{{x}(x-1)}=\dfrac {x^2+1}{x^2-x}

    After long dividing it, you're going to get this:

    \large 1 + \frac {x+1}{x(x-1)}

    Then:

    \frac {x+1}{x(x-1)} = \frac {\large A}{x} + \frac {\large B}{x-1}

    \large A(x-1)+\large B(x)= x + 1

    A = -1, B=2

    Now looks like :

    \large 1 - \dfrac {1}{x} + \dfrac {2}{x-1}
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    (Original post by sunshinejessie)
    For your first question, if you add and take away an x from the top of the equation (obviously having no overall effect) then you should be able to see that you can factorise a bit to obtain an x(x-1), which can be separated and cancelled to the 1+ that you require.

    Secondly, you're right with the idea for A and B. Do this, multiply both sides by the bottom of your fraction and compare coefficients to get values for A and B

    Hope this helps
    So what you're saying is? You can re-write the top line as x^2+1+x-x then re-write it as x(x-1)+1+x. Then splitting it into two fractions you get x(x-1)/x(x-1) + x+1/x(x-1).

    That then simplifies to 1+ (x+1)/x(x-1). I think
 
 
 
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