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We have to calculate the following integral quite often with Gauss surfaces in my Physics course. The teacher never exlained how we can find it and always gives us the result . How can I evaluate it?

$$\int \frac{dx}{(a^2+x^2)^{3/2}}$$ a is a constant

P.S. In case it's been uploaded I'll delete the question. I couldn't find it after giving a look though.

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    Did you try anything? Substituting? – 2017-01-27
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    All that comes to mind is the arctanx derivative. I don't know what substitution to try. I remember trying to solve it for quite some time back when I first saw it without luck. – 2017-01-27

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If you make the substitution $x=a\tan\theta$, then $a^2+x^2=a^2\sec^2\theta$ and $\frac{dx}{d\theta}=a\sec^2\theta$, so the result is $$ \int\frac{a\sec^2\theta}{a^3\sec^3\theta}\;d\theta=a^{-2}\int \cos\theta\;d\theta=a^{-2}\sin\theta+C=\frac{x}{a^2\sqrt{a^2+x^2}}+C$$ Note that $\sin\theta$ can be determined using $\tan\theta=\frac{x}{a}$ and the Pythagorean theorem.

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    That's very helpful, thanks! One more thing if I may. I find that many integrals require some unusual(to me at least) substitions. Are there ways to learn and recognise what I should substitute or is it just experience that helps? – 2017-01-27
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    It's mostly experience, though this particular integral is an example of a trig substitution: https://en.wikipedia.org/wiki/Trigonometric_substitution – 2017-01-27
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    @JohnKatsantas : For integrals involving $dx$ (not $da$): $$ \begin{align} \text{For } a^2 + x^2, & \text{ use } x=a\tan\theta \text{ and } 1+\tan^2\theta = \sec^2\theta. \\ \text{For } a^2-x^2, & \text{ use } x=a\cos\theta \text{ and } 1-\cos^2\theta = \sin^2\theta \\ & \text{ or } x = a\sin\theta \text{ and } 1 - \sin^2\theta = \cos^2\theta. \\ \text{For } x^2-a^2, & \text{ use } x = a\sec^2\theta \text{ and } \sec^2\theta-1 = \tan^2\theta. \end{align} $$ – 2017-01-27
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    @JohnKatsantas : In each case you use one of those trigonometric identities, so you need to see which trigonometric identity it looks like. – 2017-01-27
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    @JohnKatsantas : HOWEVER: Don't use a trigonometric substitution for things like this: $$\int \frac{x\,dx}{(a^2+x^2)^{3/2}}.$$ For that, use the substitution suggested by this: $$ \int \frac 1 {(a^2+x^2)^{3/2}} \Big( x\, dx\Big), $$ in which $x\,dx$ will become $\dfrac 1 2 \,du.$ I.e. use $u=a^2+x^2.\qquad$ – 2017-01-27
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    That's exactly what they do in the exams. They teach us substitutions like the ones you wrote above and then they give us something like this to evaluate and it never works out . It seems I have work to do. – 2017-01-27
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In the first place, "solve" is the wrong word. You are evaluating an integral. One evaluates expressions; one solves equations; one solves problems.

If you see $(a^2+x^2)^{3/2}$ and don't think of the trigonometric substitution $x = a \tan\theta$ then you need to review trigonometric substitutions. Then you have $a^2+x^2 = a^2 (1 + \tan^2\theta) = a^2 \sec^2\theta,$ so $(a^2+x^2)^{3/2} = a^3 \sec^3\theta,$ and $dx = a\sec\theta\tan\theta\,d\theta.$ So you have $$ \int \frac{dx}{(a^2+x^2)^{3/2}} = \int \frac{a\sec^2\theta\,d\theta}{a^3\sec^3\theta} = \frac 1 {a^2} \int \cos\theta\,d\theta = \cdots\cdots. $$ Once you get a function of $\theta$ you need to convert it back into a function of $x$, and that will also require remembering some trigonometry.

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    Sorry about that,I've just started learning Math terminology in English, it will take some time. Thanks for letting me know. Our teachers take some things for granted in the electrical engineer department and they never taught us these substitutions. However they always require us to evaluate(!) integrals like this and many more we've never come across. I plan to fill the gap myself after the exams, it's become very frustrating. – 2017-01-27
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    @JohnKatsantas : That is actually a frequent error among non-mathematicians. – 2017-01-27