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I had to integrate $$\int\frac{x^2+1}{(x^2-1)^2} dx$$

Well my first approach was to write$\ (x^2+1)$ as $\ (x^2-1)+2$ so as to obtain fractions $$\frac{1}{(x^2-1)} + \frac{2}{(x^2-1)^2}$$

Now I know how to integrate the first part but how to integrate the second part i.e. a quartic (biquadratic) in the denominator?

(I got the answer to the original integral by dividing num and denom by $x^2$ and then substitution,but I want to know what could I have done with my first attempt)

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    $\begingroup$ Try to find the partial fractions decomposition of your function. $\endgroup$
    – Tintarn
    Sep 11 '15 at 17:58
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The original function is already very nice. It is equal to $$\frac{1}{2}\left(\frac{1}{x^2-2x+1}+\frac{1}{x^2+2x+1}\right).$$ Now integrate.

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  • $\begingroup$ Do you see that with or without doing partial fraction decomposition? $\endgroup$
    – mickep
    Sep 11 '15 at 18:14
  • $\begingroup$ Yes, it is by inspection. Partial fractions sounds like work. $\endgroup$ Sep 11 '15 at 18:19
  • $\begingroup$ In fact, the original integral is easily solved by inspection. $\endgroup$
    – mickep
    Sep 11 '15 at 18:20
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We can Write $\displaystyle \frac{1}{(x^2-1)^2} = -\frac{1}{2}\left[\frac{1}{x^2-1}-\frac{(x^2+1)}{(x^2-1)^2}\right]$

So $$\displaystyle I = \int\frac{1}{(x^2-1)^2}dx = -\frac{1}{2}\left[\int\frac{1}{x^2-1}dx-\int\frac{x^2+1}{(x^2-1)^2}dx \right]$$

So we get $$\displaystyle I = \int\frac{1}{(x^2-1)^2}dx = -\frac{1}{4}\int\left(\frac{1}{x-1}-\frac{1}{x+1}\right)+\frac{1}{2}\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2}dx $$

Now Put $\displaystyle \left(x-\frac{1}{x}\right)=t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$

So we get $$\displaystyle \displaystyle I = -\frac{1}{4}\left[\ln|x-1|-\ln|x+1|\right]-\frac{1}{2}\cdot \frac{1}{x-\frac{1}{x}}+\mathcal{C}$$

So we get $$\displaystyle I = \int\frac{1}{(x^2-1)^2}dx = \frac{1}{4}\ln\left|\frac{x+1}{x-1}\right|-\frac{x}{2(x^2-1)}+\mathcal{C}$$

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  • $\begingroup$ It is nice that you, with your first calculations get back the original integrand. It is even nicer that you write it as $(1+1/x^2)/(x-1/x)^2$ and find a good substitution solving that integral. $\endgroup$
    – mickep
    Sep 11 '15 at 18:17
  • $\begingroup$ Thanks mickep.... I have Using same method here....math.stackexchange.com/questions/1406034/integrating-frac1x4-12/… $\endgroup$
    – juantheron
    Sep 11 '15 at 18:19
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Find the partial fraction decomposition

$$\frac 1{(x^2-1)^2}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}+\frac D{(x+1)^2}.$$

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Hint: Write is as

$$\frac{1}{(x^2-1)^2} = \frac{1}{[(x-1)(x+1)]^2} = \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2}$$

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