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The problem is:

$$\int\frac{1}{(x^2-1)^2}dx$$

I tried using substitution $u=x^2-1$ but that does not bring me got results. I get an integral:

$$\frac{1}{2}\int\frac{1}{u^2\sqrt{u+1}}$$

And that does not really make things any more simple. From here I tried using partial decomposition but didn't really get anywhere.

Any help with this would be much appreciated.

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$$\begin{align*} \int \frac{1}{(x^2 - 1)^2}\,\mathrm{d}x &\equiv \int \frac{1}{\left((x + 1)\cdot(x - 1)\right)^2}\,\mathrm{d}x \\ &= \int \frac{1}{(x + 1)^2\cdot(x - 1)^2}\,\mathrm{d}x\tag{1} \end{align*}$$ Decompose $(1)$ into partial fractions, $$\int \frac{1}{(x - 1)^2\cdot(x + 1)^2}\,\mathrm{d}x = \frac 14\int \frac{1}{x + 1} + \frac{1}{(x + 1)^2} - \frac{1}{x - 1} + \frac{1}{(x - 1)^2}\,\mathrm{d}x\tag{2}$$ Apply linearity to split the integral and then use the substitution $u = x + 1$ for the first two terms and $u = x - 1$ for the last two terms.

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Hint...you need to express this in partial fractions, i.e.$$\frac 14\left(-\frac{1}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{x+1}+\frac{1}{(x+1)^2}\right)$$

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The standard way consists in decomposing the integrand into simple fractions: $$\frac1{(x^2-1)^2}=\frac1{(x-1)^2(x+1)^2}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}+\frac D{(x+1)^2}.$$ Noting the integrand is an even function, we deduce that $C=-A$ and $D=B$.

Now multiply both sides by $(x^2-1)^2$: we obtain $$1= A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2,$$ and setting $x=1$ yields the relation $\;1=4B$, whence $B=D=\frac14$.

Next set $x=0$: we now get $1=-A+B+C+D=-2A+\frac12$, whence $A=\frac14$, $C=-\frac14$.

Can you proceed from here?

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  • $\begingroup$ Good luck for 100k. $\endgroup$ Mar 17 '18 at 23:36
  • $\begingroup$ @Salahamam_Fatima: Thanks! $\endgroup$
    – Bernard
    Mar 17 '18 at 23:41
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hint

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

$$\frac {-1}{x^2-1}+\frac {1}{2}\frac {2x\cdot x}{(x^2-1)^2} $$ use by parts for the last integrale.

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$$\int\frac{1}{(x^2-1)^2}\,dx$$

Let $ x=\sec(t)$ and $dx=\sec(t) \tan(t) \, dt$

With $$ x^2-1 = \sec^2(t) -1 = \tan^2(t) $$

The integral will change to $$\int \cot^2 (t) \, dt$$ Which is easy to solve.

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$$I=\int\frac{1}{(x^2-1)^2}\,dx$$ note that $(x^2-1)=(x-1)(x+1)$ $$I=\frac 12\int\frac{(x+1) -(x-1)}{(x^2-1)^2}\,dx$$

Simplify and do that again till you get simple fractions

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    $\begingroup$ Should be $(x^2-1)=(x-1)(x+1)$. $\endgroup$ Mar 17 '18 at 23:20
  • $\begingroup$ Yes of course @MathewMahindaratne thanks $\endgroup$
    – MtGlasser
    Mar 17 '18 at 23:23
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Hint:

Write the numerator $1$ as $\dfrac{(x+1-(x-1))^2}4$ and expand

and use $\dfrac2{(x+1)(x-1)}=\dfrac{x+1-(x-1)}{(x+1)(x-1)}$

Alternatively, integrate by parts,

$$\int \dfrac1x \cdot\dfrac x{(x^2-1)^2}dx=?$$

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