Calculate an integral of an irrational function

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!

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

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With $\ds{\quad x \equiv {t^{2} + a \over 2t}\quad}$ and $\ds{\quad t = x - \root{x^{2} - a}}$:
At this point, the integration is an elementary one. What is the relation between $\ds{a\ \mbox{and}\ b}$ ?.
• @MathsSurvivor $\displaystyle\Large \left({\bullet \qquad \bullet \atop \mid} \atop \mbox{You're welcome}\right)$ Sep 6 '18 at 15:44