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This question already has an answer here:

How can I evaluate the following integral

$$\int {\frac{1}{{4 - 9{x^2}}}dx} $$

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marked as duplicate by Did, Travis Willse, N. F. Taussig, user147263, user251257 Nov 1 '15 at 19:15

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    $\begingroup$ It's $\int \frac{dx}{(2+3x)(2-3x)}$, then use Partial Fraction decomposition. $\endgroup$ – user236182 Nov 1 '15 at 13:17
  • $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution.Consider accepting an useful answe. You could have looked up on Wolfram Alpha $\endgroup$ – Shailesh Nov 1 '15 at 13:25
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Notice, use partial fractions as follows $$\int \frac{dx}{4-9x^2}=\int \frac{dx}{(2-3x)(2+3x)}$$ $$=\int \frac{1}{4}\left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$ $$=\frac{1}{4}\int \left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$

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Hint:

The derivative of $\tanh^{-1}(x)$ is $$\frac 1{1-x^2}.$$ Then the integral will be of the form $$a\tanh^{-1}(bx).$$

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Isn't it this term reminding arctanh derivative ? :-) for $\frac{1}{a+b x^2}$ -like integrands, first re-check $arctan(ax)$ and $arctanh(ax)$ derivatives, choose the best appropriate, then fit to your numbers.

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  • $\begingroup$ It's confusing to mention $\frac{1}{a+bx^2}$ and then $\arctan(ax)$, when the real point is that $\frac{d}{dx}\arctan(ax)=\frac{a}{1-a^2x^2}$. No need for $\arctan$, just use implicit differentiation on $\tanh y=ax$ and use the identity $\tanh^2 y+\mathrm{sech}^2 y=1$. I don't think the answer deserves two downvotes, though, as it does suggest a viable approach. +1 $\endgroup$ – Marconius Nov 1 '15 at 14:18
  • $\begingroup$ Who deleted this answer, and why ??? $\endgroup$ – Fabrice NEYRET Nov 1 '15 at 16:17
  • $\begingroup$ I guess we are here to give some useful knowledge and generalities, not just all-cooked solutions. I do think telling how to address the just a bit more general pattern of $\frac{1}{a+bx^2}$ is an appropriate way to answer. I've cpolished it a bit, though. $\endgroup$ – Fabrice NEYRET Nov 1 '15 at 16:20
  • $\begingroup$ The post was deleted from the review queue (it was automatically put in the queue since it was so short). Apparently, the reviewers considered the original version to be not an answer. Make your answers a bit longer to avoid such things. $\endgroup$ – Daniel Fischer Nov 1 '15 at 16:43
  • $\begingroup$ Thanks for the explaination. Deletion without explaination feels extremely rude; isn't it at least possible to send a private message in such case ? $\endgroup$ – Fabrice NEYRET Nov 1 '15 at 16:56

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