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How can I calculate this integral

$$\int \frac{dx}{(x^2+b)\sqrt{x^2-a}} $$

Without using the substitution $$x=\sqrt{a}\sec {u} $$

I guess I'm having some problems while using secant substitution so I wanted to know if there is any other possible way to solve the above integral . Any hint will be appreciated. Thank you in return!

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  • $\begingroup$ Mathematica gives: $\frac{\tan ^{-1}\left(\frac{x \sqrt{-a-b}}{\sqrt{b} \sqrt{x^2-a}}\right)}{\sqrt{b} \sqrt{-a-b}}$. $\endgroup$ Sep 4 '18 at 16:09
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    $\begingroup$ @DavidG.Stork how? $\endgroup$ Sep 4 '18 at 16:11
  • $\begingroup$ Through the substitution $x=\sqrt{a} \sec u$. $\endgroup$ Sep 4 '18 at 16:25
  • $\begingroup$ But I wanted another way of solving it without using this substitution $\endgroup$ Sep 4 '18 at 16:29
  • $\begingroup$ Don't know another method. $\endgroup$ Sep 4 '18 at 16:31
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With $\ds{\quad x \equiv {t^{2} + a \over 2t}\quad}$ and $\ds{\quad t = x - \root{x^{2} - a}}$:

\begin{align} &\bbox[10px,#ffd]{\ds{\int{\dd x \over \pars{x^{2} + b}\root{x^{2} - a}}}} = -\int{4t \over t^{4} + 2\pars{a + 2b}t^{2} + a^{2}}\,\dd t \\[5mm] \stackrel{y\ =\ t^{2}}{=}\,\,\,& -2\int{\dd y \over y^{2} + 2\pars{a + 2b}y + a^{2}} \\[5mm] = &\ -2\int{\dd y \over \pars{y + a + 2b}^{2} + a^{2} - \pars{a + 2b}^{2}} = -2\int{\dd y \over \pars{y + a + 2b}^{2} - 4b\pars{b - a}} \end{align}

At this point, the integration is an elementary one. What is the relation between $\ds{a\ \mbox{and}\ b}$ ?.

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  • $\begingroup$ a and b are just random numbers , there's no specific relation between them. $\endgroup$ Sep 6 '18 at 10:34
  • $\begingroup$ @MathsSurvivor 0 k. It's fine. $\endgroup$ Sep 6 '18 at 14:30
  • $\begingroup$ thank you for your answer! $\endgroup$ Sep 6 '18 at 14:39
  • $\begingroup$ @MathsSurvivor $\displaystyle\Large \left({\bullet \qquad \bullet \atop \mid} \atop \mbox{You're welcome}\right)$ $\endgroup$ Sep 6 '18 at 15:44

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