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I was having a bit of trouble with an integral

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

It was given as a partial fractions problem, and I tried a substitution of

$x=sec(\theta);dx=sec(\theta)tan(\theta)d\theta$

to get $$\int{{\sec(\theta) \tan(\theta)}\over{\tan^4(\theta)}}d\theta=\int{\cos^2(\theta)\over \sin^3(\theta)}d\theta$$

but I can't seem to move past this. Any suggestions would be greatly appreciated.

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  • $\begingroup$ No, and I'm not quite sure how you did get that. I'm sorry, I'm really new to this. $\endgroup$ – user218279 Feb 22 '15 at 3:50
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We break up the given integral into four parts by a partial fraction decomposition and get four elementary integrals: $$ \begin{eqnarray*} \int\frac{1}{(x^2-1)^2}dx & = & \int\frac{1}{(x+1)^2(x-1)^2}dx \\ & = &\int\frac{1}{4}\left(\frac{1}{(x+1)^2}+ \frac{1}{(x-1)^2}+ \frac{1}{x-1} + \frac{1}{x+1}\right)dx \\ & =& \frac{1}{4}\int\frac{1}{(x+1)^2}dx + \frac{1}{4}\int\frac{1}{(x-1)^2}dx + \frac{1}{4}\int\frac{1}{x-1}dx + \frac{1}{4}\int\frac{1}{x+1}dx \\ & \stackrel{[u = x + 1]}{=} & \frac{1}{4}\int\frac{1}{u^2}du + \frac{1}{4}\int\frac{1}{(x-1)^2}dx + \frac{1}{4}\int\frac{1}{x-1}dx + \frac{1}{4}\int\frac{1}{u}du \\ & = & -\frac{1}{4u} + \frac{1}{4}\int\frac{1}{(x-1)^2}dx + \frac{1}{4}\int\frac{1}{x-1}dx + \frac{1}{4}\ln(u) \\ & \stackrel{[v = x - 1]}{=} & -\frac{1}{4u} + \frac{1}{4}\int\frac{1}{v^2}dv + \frac{1}{4}\int\frac{1}{v}dv + \frac{1}{4}\ln(u) \\ & = & -\frac{1}{4u} - \frac{1}{4v} - \frac{1}{4}\ln(v) + \frac{1}{4}\ln(u) \\ & = & -\frac{1}{4(x+1)} - \frac{1}{4(x-1)} -\frac{1}{4}\ln(x-1) + \frac{1}{4}\ln(x+1) \\ & = &\frac{1}{4}\left(-\frac{2x}{x^2-1}-\ln(1-x) + \ln(x+1)\right). \end{eqnarray*} $$

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Break the integral up into manageable parts. \begin{align*} \int \frac{1}{(x^2-1)^2}dx &= \int \frac{1}{2(1-x^2)}dx + \int\frac{x^2+1}{2(x^2-1)^2}dx \\&= \frac{1}{2} \text{tanh}^{-1}{x} \hspace{1mm} + \int\frac{x^2+1}{2(x^2-1)^2}dx \\ &= \frac{1}{2} \text{tanh}^{-1}{x} \hspace{1mm} - \frac{x}{2(x^2-1)}\hspace{1mm}+\text{C}\\ &=\frac{1}{2}(\text{tanh}^{-1}{x}\hspace{1mm} - \frac{x}{x^2-1}) + \text{C} \end{align*}

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You could proceed by writing $\displaystyle \int\frac{\cos^2\theta}{\sin^3\theta}d\theta=\int\cot^2\theta\csc\theta\;d\theta$;

now use integration by parts with $u=\cot\theta, dv=\cot\theta\csc\theta$ and

then use the identity $\csc^2\theta=\cot^2\theta+1$.


Alternatively, you can use partial fractions with $\displaystyle\frac{1}{(x^2-1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}.$

Solving $1=A(x+1)(x-1)^2+B(x-1)^2+C(x+1)^2(x-1)+D(x+1)^2$

gives $A=\frac{1}{4}, B=\frac{1}{4}, C=-\frac{1}{4}, D=\frac{1}{4}$, and now just integrate each term as in Maria's answer.

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