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What method should be used to determine the antiderivative of this expression?

Edit: I have $ c > 0 $ in the problem I'm working on.

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closed as off-topic by GEdgar, abiessu, Claude Leibovici, user99914, kjetil b halvorsen Apr 21 '15 at 6:50

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    $\begingroup$ For the usual approaches, it makes a difference whether $c\lt 0$, $c=0$, or $c\gt 0$. $\endgroup$ – André Nicolas Apr 21 '15 at 3:14
  • $\begingroup$ @AndréNicolas So, we can't evaluate this generally for any $c$? $\endgroup$ – Al Jebr Apr 21 '15 at 3:16
  • $\begingroup$ We can. But the standard trigonometric substitutions are different for $c\gt 0$ and for $c\lt 0$. $\endgroup$ – André Nicolas Apr 21 '15 at 4:06
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Try trigonmetric substitution.

Examples can be found here: http://en.wikipedia.org/wiki/Trigonometric_substitution

In particular, try the substitution $x=\sqrt{c}\tan\theta$ if $c > 0$. Try $x=\sqrt{|c|}\sec\theta$ if $c < 0$

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$$\int(x^2+c)^{-3/2}dx=\int\frac{1}{\sqrt{(x^2+c)^3}}$$

Let us use a new variable $t$ such that $\sqrt{x^2+c}=x+t$. This is the first Euler substitution.

From this formula we can solve for $x$ to get $x=\frac{c-t^2}{2t}$, $dx=\frac{-c-t^2}{2t^2}dt$, $\sqrt{x^2+c}=x+t=\frac{c+t^2}{2t}$, and $\sqrt{(x^2+c)^3}=\frac{(c+t^2)^3}{8t^3}$.

We put these into the integral and get

$$\int\frac{8t^3}{(c+t^2)^3}\frac{-(c+t^2)}{2t^2}dt=-2\int\frac{2t}{(c+t^2)^2}dt=2\frac{1}{c+t^2}$$

We can return to the variable $x$ since $t=\sqrt{x^2+c}-x$. Therefore

$$\int(x^2+c)^{-3/2}=\frac{2}{c+\left(\sqrt{x^2+c}-x\right)^2}$$

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