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Good day! I'm currently having an exercise on partial fractions... I know the basic of different methods in separating a fraction into smaller parts but I got confused when I encountered this one$$\frac {(n + 1)^2 (t^{2n})}{(t^{2n+2} - 1)^{2}}$$ is there any way on how to deal with this 2 variable fraction? I tried wolfram online calculator and I got this $$\frac{n^2}{(4t^2 (t^{n+1}-1) )}- \frac {n^2}{(4t^2 (t^{n+1}+1) )}+ \frac{n^2}{(4t^2 (t^{n+1}-1)^2 )}+ \frac {n^2}{(4t^2 (t^{n+1}+1)^2 )}+ \frac{n}{(2t^2 (t^{n+1}-1) )}- \frac{n}{(2t^2 (t^{n+1}+1) )}+ \frac{n}{(2t^2 (t^{n+1}-1)^2 )}+ \frac {n}{(2t^2 (t^{n+1}-1)^2 )}+ \frac {1}{(4t^2 (t^{n+1}-1) )}- \frac{1}{(4t^2 (t^{n+1}+1) )}+ \frac{1}{(4t^2 (t^{n+1}-1)^2 )}+ \frac {1}{(4t^2 (t^{n+1}+1)^2 }$$

But I'm not onlu interested to the answer alone but to the solution itself... any idea would be a great help... thanks!

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You want to re-express $x$ in

$x = \displaystyle\frac {(n + 1)^2 (t^{2n})}{(t^{2n+2} - 1)^{2}}$

Write $x = \dfrac{(n + 1)^2}{t^2} \cdot \dfrac{t^{2n+2}}{(t^{2n+2} - 1)^{2}}$

Then put $u = t^{n+1}$ to get $x = \dfrac{(n + 1)^2}{t^2} \cdot \dfrac{u^2}{(u^2 - 1)^{2}} \tag{1}$

and then because we are dealing with repeated factors

$\dfrac{u^2}{(u^2 - 1)^2} = \dfrac{u^2}{(u+1)^2(u-1)^2} = \dfrac{A}{u+1} + \dfrac{B}{(u+1)^2} + \dfrac{C}{u-1} + \dfrac{D}{(u-1)^2} \tag{2}$.

Then multiply the last equation through by the full denominator to get

$u^2 = A(u+1)(u-1)^2+B(u-1)^2+C(u+1)^2(u-1)+D(u+1)^2 \tag{3}$

Now evaluate at $u=-1, u=1$ (some terms are zero) to get the two equations

$\begin{align} 1 &= 4B \\ 1 &= 4D \end{align}$

whence $\boxed{B=D=\frac{1}{4}}$.

Taking the derivative of (3) wrt $u$:

$2u = A[(u-1)^2+2(u+1)(u-1)]+2B(u-1)+C[2(u+1)(u-1)+(u+1)^2]+2D(u+1) \tag{4}$

Evaluate at $u=-1, u=1$ to get the two equations:

$\begin{align} -2 &= 4A - 4B &= 4A - 1 \\ 2 &= 4C + 4D &= 4C + 1 \end{align}$

whence $\boxed{A=-\frac{1}{4}, C=\frac{1}{4}}$

Therefore

$\dfrac{u^2}{(u^2 - 1)^2} = \dfrac{1}{4} \Bigg\{\cdot \dfrac{-1}{u+1} + \dfrac{1}{(u+1)^2} + \dfrac{1}{u-1} + \dfrac{1}{(u-1)^2}\Bigg\} \tag{5}$

Now substitute (5) into (1) then apply $u=t^{n+1}$ to get the complete expression for $x$.

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  • $\begingroup$ thank a lot @Marconius taking the derivative does not come into my mind... tnx a lot! $\endgroup$ – rosa Jul 13 '15 at 15:10
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i think:

$ (n+1)^2 \frac{t^{2n}}{(t^{n+1}-1)(t^{n+1}+1)} $

$ u=t^n $

$ (n+1)^2 \frac{u^2}{(u^2-1)^2(u^2+1)^2} $

now, first decompose this , and then return to original variable, t, and decompose again the rest of terms

or again change the variable as below:

$ z=u^2 $

and then do the aforementioned sequential decomposition!

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