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This was in an old exam in a physics for mathematicians class. I haven't had to deal with these kind of integrals for a while and can't think of a decent substitution. I asked my teacher about it and he mumbled a bit and told me to "google it".

I've tried a few obvious ones but none of them seemed to work.

Any help would be greatly appreciated.

John Williams
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4 Answers4

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Integration by parts easily gives: $$\int\frac{dx}{(x^2+A)^{3/2}} = C+\frac{x}{A\sqrt{x^2+A}}.$$

Jack D'Aurizio
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    and the parts are ? – d_e Jun 26 '15 at 20:56
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    @d_e: what do you think? – Jack D'Aurizio Jun 26 '15 at 20:57
  • @JackD'Aurizio What is with all of the negative comments left the past few days by others? +1 I don't understand how others take some pleasure in unduly criticizing others who are earnestly trying to help. – Mark Viola Jun 26 '15 at 21:33
  • @Dr.MV: not every top user on MSE has a good temper (maybe, neither do I). If you plan to become one of them (I guess so), you have to be prepared to fight an opinion war, from time to time. – Jack D'Aurizio Jun 26 '15 at 21:42
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    @JackD'Aurizio I posted an answer the other day, it was correct, but I omitted a step. I was summarily lambasted for writing something that was "not correct" (it was correct as were the intermediate parts - just not obvious to a top user). 23 comments ensued until I edited with the "non-obvious" steps. So, I understand your comment. But there are a few on here that just seem to like to impolitely criticize. – Mark Viola Jun 26 '15 at 21:49
  • just have to wonder if you were responding to my first comment here accusing me of having "bad temper" or being negative. just let me say, it has nothing to do with "critics" and I am sorry if it was understood that way. MSE was different place without people like you really helping. – d_e Jun 27 '15 at 19:26
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Classic Substitution Method

Let $x=\sqrt{y^2+z^2}\tan \theta$ and thus $dx=\sqrt{y^2+z^2}\sec^2 \theta \,d\theta$. Then,

$$\begin{align} \int\frac{dx}{(x^2+y^2+z^2)^{3/2}}&=\frac{1}{y^2+z^2}\int\cos \theta \,d\theta\\\\ &=\frac{1}{y^2+z^2}\sin \theta +C\\\\ &=\frac{1}{y^2+z^2}\frac{x}{\sqrt{x^2+y^2+z^2}}+C \end{align}$$

Mark Viola
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Since $y,z$ are constant let $\alpha^{2}=y^{2}+z^{2}$. Then let $x=\alpha \sinh t$. $dx = \alpha \cosh t$, so $$\int \frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}dx=\int\frac{\alpha \cosh t}{(\alpha^{2}\sinh^{2} t + \alpha^{2})^{3/2}}dt\\=\int\frac{\cosh t}{\alpha^{2}\cosh t}dt=\frac{t}{\alpha^{2}}=\frac{1}{\alpha^{2}}\sinh^{-1} \frac{x}{\alpha}$$

preferred_anon
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Following daniel's answer..

$\int \frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}dx$ With $a^2 = y^2 + z^2$

$ \therefore \int \frac{1}{(x^{2}+a^{2})^{3/2}}dx$

Hyperbolic substitutions $x=a\dot{}sinht$ and $dx=a\dot{}cosht$

$\int\frac{{a\dot{}cosh(t)}}{(a^2\dot{}sinh^{2}(t)+a^2)^{3/2}}dt$

$cosh^{2}(t)-sinh^{2}(t)=1$

$a^{2}(cosh^2(t)) = a^2(1+sinh^{2}(t)) = a^2cosh^2(t)=a^2sinh^2(t)+a^2$

Which becomes $\int\frac{{a\dot{}cosh(t)}}{(a^2\dot{}cosh^{2}(t))^{3/2}}dt=\int\frac{{(a\dot{}cosh(t))}}{(a^3\dot{}cosh^{3}(t))}dt=\int\frac{{(a\dot{}cosh(t))}}{(a\dot{}cosh(t))^{3}}dt=\int\frac{{dt}}{(a\dot{}cosh(t))^2}$

Of course I might be wrong but, I think you did something wrong with the exponential fraction part. And as it stands, This is probably even harder than the original question.

I would have loved to comment instead. Apparently I'm not allowed. The answer by Jack is correct.

Gixbit
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