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how does one integrate: $$\int \frac{1}{x^2-a^2}dx$$

I know it looks very similar to the known formula $$\int \frac{1}{x^2+a^2}dx$$ but it doesn't help really.

note: Im not allowed to use the partial fraction method in the solution here.

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    $\begingroup$ You say you are not allowed to use partial fractions, yet you accept such answer? O_o $\endgroup$ – user2345215 Jan 3 '15 at 11:51
  • $\begingroup$ @user2345215 Partial fraction decomposition $\endgroup$ – Bak1139 Jan 3 '15 at 11:54
  • $\begingroup$ @Bak1139 Partial fractions and partial fraction decomposition are the same things. $\endgroup$ – Ali Caglayan Jan 3 '15 at 13:47
  • $\begingroup$ Hey man. I am going to delete my answer so accept the other method :). I don't think this question deserves a down vote so you get a +1 from me. $\endgroup$ – Chinny84 Jan 3 '15 at 14:07
  • $\begingroup$ ok sure...still a good way though. $\endgroup$ – Bak1139 Jan 3 '15 at 14:28
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If you substitute $x=a\tanh y$ you get $$\int \frac{1}{a^2\tanh^2y-a^2}a\text{sech}^2 ydy$$

This is similar to that $\tan$ substitution you might use for your known formula

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  • $\begingroup$ This is the answer that should be accepted. The currently accepted answer uses partial fractions which the OP specifically asked not to use. Yet he accepted a partial fractions answer... +1 for this though. I would also provide a link to somewhere explaining hyperbolic functions if the OP is unfamiliar with them. $\endgroup$ – Ali Caglayan Jan 3 '15 at 13:46
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Hint: $$\frac{1}{x^2-a^2} = \frac{1}{(x-a)(x + a)} = \frac{A}{x-a} + \frac{B}{x+a}$$

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  • $\begingroup$ unfortunately im not allowed to use the fraction method $\endgroup$ – Bak1139 Jan 3 '15 at 11:38
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    $\begingroup$ Well, you've accepted Chinny's answer... $\endgroup$ – Aaron Maroja Jan 3 '15 at 11:51
  • $\begingroup$ i meant disintegration into a and b as in Partial fraction decomposition $\endgroup$ – Bak1139 Jan 3 '15 at 11:53
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    $\begingroup$ That's exactly what he/she did. $\endgroup$ – Aaron Maroja Jan 3 '15 at 11:54
  • $\begingroup$ well I thought the two were different. $\endgroup$ – Bak1139 Jan 3 '15 at 11:58
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You can use :

$$\int \frac{1}{x^2 - a^2}dx = \int \frac{-1}{a^2} \frac{1}{1 - \frac{x^2}{a^2} }dx$$

Which is known, then $$\int \frac{1}{x^2 - a^2}dx = \frac{-\text{arctanh}(\frac{x}{a})} {a}$$

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Note that

$${a\over x^2-a^2}={x+a-x\over x^2-a^2}={1\over x-a}-{x\over x^2-a^2}$$

Therefore

$$\begin{align} \int{1\over x^2-a^2}dx&={1\over a}\int{a\over x^2-a^2}dx\\ \\ &={1\over a}\int{1\over x-a}dx-{1\over a}\int{x\over x^2-a^2}dx\\ \\ &={1\over a}\ln|x-a|-{1\over2a}\ln|x^2-a^2|+C \end{align}$$

Remark: This looks a bit (or a lot) like partial fractions, but it's not the standard partial fraction method.

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Well from the Pythagorean Identities of trig we have:

$\tan^{2}(\theta) + 1 = \sec^{2}(\theta)$

this tells us that if we let $x=a\sec\theta$ we can then reduce this to solving

$$\int\frac{a\sec\theta\tan\theta}{a^2\sec^2\theta-a^2}d\theta$$

I'm assuming you can take it from here.

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