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According to the Wolfram integrator: $$\int (x^2+a)^{-3/2}\text{d}x = {x \over a \times \sqrt{(x^2+a)}}$$

I easily differentiated the answer to verify that it was correct (not that I don't trust Wolfram or anything), but how would I get this result on my own? I'm reasonably sure that integration by parts should be used, although I tried splitting it up a few different ways and I couldn't get it to work. A short hint would probably suffice. (You might be able to guess what physics problem I was working on when I encountered this integral).

graydad
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pmennen
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    Trig substitution. $x=\sqrt a \cdot \tan \theta$. – Christopher Carl Heckman Nov 19 '15 at 05:26
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    You can also try $x=\sqrt a \sinh(y)$ – Claude Leibovici Nov 19 '15 at 05:41
  • This integral isn't unique to any particular physics phenomena... but I'll guess for fun. An E&M problem out of Griffiths? – zahbaz Nov 19 '15 at 05:49
  • Those $3/2$ exponents, lol. Never seen one outside of EM. – Benjamin Lindqvist Nov 19 '15 at 06:53
  • Correct zahbaz, it was E&M. I was solving for the magnetic field of an infinite wire from the Biot Savart law, although the same integral is needed for the E field of an infinite wire from Columbs law. I probably should have written a^2 instead of a in my question since that is how it arose in the problem. A changed it to a because it made the integral and the result look simpler, but now I see that the substitution solution looks simpler with the a^2. Thanks for all the answers! – pmennen Nov 20 '15 at 15:06

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HINT:

Assuming all variables are postitive:

$$\int\left(x^2+a\right)^{-\frac{3}{2}}\space\text{d}x=$$ $$\int\frac{1}{\left(x^2+a\right)^{\frac{3}{2}}}\space\text{d}x=$$

Substitute $x=\sqrt{a}$ and $\text{d}x=\sqrt{a}\sec^2(u)\space\text{d}u$. Then $\left(x^2+a\right)^{\frac{3}{2}}=\left(a\tan^2(u)+a\right)^{\frac{3}{2}}=a^{\frac{3}{2}}\sec^3(u)$ and $u=\arctan\left(\frac{x}{\sqrt{a}}\right)$:

$$\sqrt{a}\int\frac{\cos(u)}{a^{\frac{3}{2}}}\space\text{d}u=$$ $$\frac{1}{a}\int\cos(u)\space\text{d}u$$

Jan Eerland
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