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

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

This came up doing a physics task, but I have no idea how to integrate it without straight using integration table. I tried to do it integrating by parts but that get me nowhere.

• Inverse square law forces? Sep 2 '14 at 14:11
• electric field near an infinitely long line charge
– Lugi
Sep 2 '14 at 14:13
• Why don't you use Gauss' LAw? Sep 2 '14 at 14:13
• Because the taks was to do it using Couloumb's law
– Lugi
Sep 2 '14 at 14:16

A start: Let $x=R\tan\theta$. After a while, you will be integrating $\cos\theta$.

More: Then $dx=R\sec^2\theta \,d\theta$. Note that $$R^2+x^2=R^2+R^2\tan^2\theta=R^2(1+\tan^2\theta)=R^2\sec^2\theta.$$ It follows that $(R^2+x^2)^{3/2}=R^3\sec^3\theta$. Thus $$\int \frac{1}{(R^2+x^2)^{3/2}}\,dx=\int \frac{R\sec^2\theta}{R^3\sec^3\theta}\,d\theta=\int \frac{1}{R^2}\cos \theta\,d\theta.$$

• Just what we physics students are taught as a rule for this repeatedly occurring integral, so much that I almost remembered the results for various limits sometime back. Sep 2 '14 at 14:13
• Because of inverse distance laws, this sort of thing does come up often. For me, it feels better to deal with an integral that is "real" than with something artificial that succombs to a "clever" trick. Sep 2 '14 at 14:27
• Im stuck at $\int \frac {1-\tan \theta}{\sqrt{1+\tan\theta }} d\theta$ . What do I do from here on?
– Lugi
Sep 2 '14 at 19:08
• I will type a few additional lines in the answer, typing formulas in a comment often gives me TeX problems. Sep 2 '14 at 20:11
• Ok, this integral equals to $\frac {\sin\theta}{R^2}$ but what should I do with it to get x involved? The best I can come up with is $\frac {x \cos\theta}{R^3}$
– Lugi
Sep 2 '14 at 20:50