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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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    Is this homework?2012-12-17

4 Answers 4

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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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    @Graphth Ty, fixed. Was just a typo, used the right one in the calculation....2012-12-17
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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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    I think you missed the "without trigonometric substitution" part...2012-12-17
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    @DonAntonio, thanks for pointing out. How about this one?2012-12-17
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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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    are hyperbolic function not cognate of trigonometric functions?2012-12-17
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    Perhaps they are @lab, but they are *not* trigonometric functions per se: they are exponential ones. BTW, did you downvote my answer?2012-12-17
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    @I have not yet downvoted in math SE2012-12-17
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    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.2012-12-17
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    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.2012-12-17
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    @labbhattacharjee: You do not even need hyperbolic functions if you remember that $2\cosh x =e^{x}+e^{-x}$.2012-12-17
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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 $u=4t^2+1$.