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How can I evaluate the following integral

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

Hosein Rahnama
  • 14,882
David
  • 1

3 Answers3

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Notice, use partial fractions as follows $$\int \frac{dx}{4-9x^2}=\int \frac{dx}{(2-3x)(2+3x)}$$ $$=\int \frac{1}{4}\left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$ $$=\frac{1}{4}\int \left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$

Harish Chandra Rajpoot
  • 37,450
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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.

Fabrice NEYRET
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  • 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 – Marconius Nov 01 '15 at 14:18
  • Who deleted this answer, and why ??? – Fabrice NEYRET Nov 01 '15 at 16:17
  • 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. – Fabrice NEYRET Nov 01 '15 at 16:20
  • 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. – Daniel Fischer Nov 01 '15 at 16:43
  • Thanks for the explaination. Deletion without explaination feels extremely rude; isn't it at least possible to send a private message in such case ? – Fabrice NEYRET Nov 01 '15 at 16:56