$$\int\frac{dx}{(z^{2}+x^{2})^{3/2}}$$ I arrived at this after a substitution $t=x/z$: $$\frac{1}{z^{2}}\int\frac{dt}{(1+t^{2})^{3/2}}$$ but now stuck with that 3/2 in the exponent.
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3Try the standard substitution $x=z \tan(t)$. – Teddy Feb 11 '13 at 10:51
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1Try a trigonometric substitution: construct a right triangle with catheti $z$ and $x$, and see what happens... – Matemáticos Chibchas Feb 11 '13 at 10:53
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Let us substitute $x = z \tan(t)$. Differentiating we get $ dx = \frac{z\, dt}{\cos^2(t)}$. Now, $$ \int \frac{z\,dt}{\cos^2(t) z^3 (1+\tan^2(t))^{3/2}}=$$ $$ \frac{1}{z^2} \int \frac{dt}{\cos^2(t) (\frac{1}{\cos^2(t)})^{3/2}}=$$ $$ \frac{1}{z^2} \int \cos(t) \,dt= \frac{1}{z^2} \sin(t) + C = $$ $$ \frac{1}{z^2} \frac{x}{\sqrt{z^2+x^2}} + C,$$ where the back-substitution is done (as in the above remark by Matemáticos Chibchas) by constructing a right triangle.
Teddy
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Formally you're right. However, $x$ runs over all real values as $t\in (-\pi/2, \pi /2)$, and in this last interval, $\cos(t)>0$. – Teddy Feb 12 '13 at 11:20
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You can also try a hyperbolic substitution:
$$\frac{x}{z}=\sinh u\Longrightarrow dx=z\cosh u$$
$$\int\frac{dx}{(z^2+x^2)^{3/2}}=\frac{1}{z^3}\int z\cosh u\; du\frac{1}{(1+\sinh^2u)^{3/2}}=\frac{1}{z^2}\int\frac{du}{\cosh^2 u}=$$
$$=\frac{2}{z^2}\int\frac{e^udu}{e^{2u}+1}=\frac{2}{z^2}\arctan e^u+C\ldots$$
DonAntonio
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