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Integrate the indefinite integral ?

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


My try :-

Let $x^2 = t \to 2xdx=dt$ and then rationalize

I have,

$$\frac{1}{4a^2}\int \sqrt{t-a^2}- \sqrt{t+a^2}dt$$


I know its silly, but I don't know how to get further ?

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  • $\begingroup$ I would multiply the given integral top and bottom with the denominator's conjugate. That way you can split the integral in two since the new denominator is only $-2a^2$ $\endgroup$ – imranfat Apr 19 '17 at 18:21
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    $\begingroup$ for your first integral se $$x=a\cosh(t)$$ and for the second one $$x=a\tan(t)$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 19 '17 at 18:21
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    $\begingroup$ we have $$x^2-a^2-x^2-a^2=-2a^2$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 19 '17 at 18:23
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For example, $$ \int(t-a^2)^{1/2}\,dt=\frac{2}{3}(t-a^2)^{3/2}+C $$

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  • $\begingroup$ Is there no need to go inside the square root after what you did and integrate ? $\endgroup$ – Jon Garrick Apr 19 '17 at 18:23
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    $\begingroup$ Jon, no because the Chain rule does not "do" anything here, derivative of $t-a^2$ with respect to $t$ is.... $\endgroup$ – imranfat Apr 19 '17 at 18:24
  • $\begingroup$ No, try to differentiate what I wrote. $\endgroup$ – mickep Apr 19 '17 at 18:24
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    $\begingroup$ Thanks @imranfat, fixed. $\endgroup$ – mickep Apr 19 '17 at 18:24
  • $\begingroup$ @mickep I don't know what was i thinking, but yeah it was silly, thanks :)) $\endgroup$ – Jon Garrick Apr 19 '17 at 18:26
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Rationalizing yields $\int\frac{x\left(\sqrt{x^2+a^2}-\sqrt{x^2-a^2}\right)}{2a^2}$ using your substitution the integral becomes $\color{red}{\frac{1}{4a^2}}\displaystyle\int\sqrt{t+a^2}-\sqrt{t-a^2}\,\mathrm{d}x$. Because the derivative of $t+$ any constant is just $1$, we simply integrate, $\frac{1}{4a^2}\left(\frac{2}{3}(t+a^2)^{3/2}-\frac{2}{3}(t-a^2)^{3/2}\right)+C$ and then substitute back $t=x^2$

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