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I have been looking for a possible solution but they are with trigonometric integration..

I need a solution for this function without trigonometric integration

$$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$$

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  • $\begingroup$ Is this homework? $\endgroup$ – GEdgar Dec 17 '12 at 18:40
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Let $I=\int \frac{dx}{\sqrt{x^2+4}} \,;\, J=\int \frac{dx}{\sqrt{x^2+4}^3}$

We integrate $I$ by parts, we get:

$u=(x^2+4)^{-1/2}, dv=dx$

$du=-x(x^2+4)^{-3/2}, v=x$

Thus

$$I= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2}{\sqrt{x^2+4}^3}= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2+4-4}{\sqrt{x^2+4}^3}=$$ $$= \frac{x}{\sqrt{x^2+4}}+ I-4J=$$

Thus, canceling the $I$ we get

$$4J= \frac{x}{\sqrt{x^2+4}} +C$$

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  • $\begingroup$ @Graphth Ty, fixed. Was just a typo, used the right one in the calculation.... $\endgroup$ – N. S. Dec 17 '12 at 19:21
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$$\int\frac{dx}{(a^2+x^2)^{\frac n2}}=\int1\cdot \frac1{(a^2+x^2)^{\frac n2}}dx$$

$$=\frac x{(a^2+x^2)^{\frac n2}}-\int\left(\frac{-n}2\frac{2x\cdot x}{(a^2+x^2)^{\frac n2+1}} \right) dx$$

$$=\frac x{(a^2+x^2)^{\frac n2}}+n\int \left(\frac{(a^2+x^2-a^2)}{(a^2+x^2)^{\frac n2+1}}\right)dx $$

$$=\frac x{(a^2+x^2)^{\frac n2}}+n\left(\int\frac{dx}{(a^2+x^2)^{\frac n2}}-a^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}\right)+c $$ where $c$ is the constant for indefinite integration.

or, $$na^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}=\frac x{(a^2+x^2)^{\frac n2}}+(n-1)\int\frac{dx}{(a^2+x^2)^{\frac n2}}+c$$

Putting $n=1,$ $$a^2\int\frac{dx}{(a^2+x^2)^{\frac 32}}=\frac x{(a^2+x^2)^{\frac 12}}+c$$

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  • $\begingroup$ I think you missed the "without trigonometric substitution" part... $\endgroup$ – DonAntonio Dec 17 '12 at 17:22
  • $\begingroup$ @DonAntonio, thanks for pointing out. How about this one? $\endgroup$ – lab bhattacharjee Dec 17 '12 at 18:17
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$$\frac{1}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\cdot\frac{1}{\left(1+\left(\frac{x}{2}\right)^2\right)^{3/2}}$$

Now try

$$x=2\sinh u\implies dx=2 \cosh u\,du\implies$$

$$\int\frac{dx}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\int\frac{2\,du\cosh u}{(1+\sinh^2u)^{3/2}}=\frac{1}{4}\int\frac{du}{\cosh^2u}=\ldots $$

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    $\begingroup$ are hyperbolic function not cognate of trigonometric functions? $\endgroup$ – lab bhattacharjee Dec 17 '12 at 17:35
  • $\begingroup$ Perhaps they are @lab, but they are not trigonometric functions per se: they are exponential ones. BTW, did you downvote my answer? $\endgroup$ – DonAntonio Dec 17 '12 at 17:51
  • $\begingroup$ @I have not yet downvoted in math SE $\endgroup$ – lab bhattacharjee Dec 17 '12 at 17:52
  • $\begingroup$ I was just clarifying my doubt.I second, down-vote should be accompanied by an explanation. But,here if trigonometric function is not allowed, so should be hyperbolic ones. $\endgroup$ – lab bhattacharjee Dec 17 '12 at 17:59
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    $\begingroup$ I think the same as you about downvotes. About the second part: I'm not sure about that. After all, the name refers to some basic properties the hyperbolics have which ressemble trigonmetric functions...but they aren't. I'm not sure why the OP didn't want trig. substitution, but it may be some personal bias against them which, again perhaps, doesn't exist abbout the hyp's. $\endgroup$ – DonAntonio Dec 17 '12 at 19:12
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The following is not quite right, it needs minor modification for negative $x$. Let $x=\dfrac{1}{t}$. Then $dx=-\dfrac{1}{t^2}$. Substitute and do some algebra. There is some nice cancellation, and we end up with $$\int \frac{-t\,dt}{(4t^2+1)^{3/2}}.$$ Now let $u=4t^2+1$.

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