Integral of inverse of square root of a quadratic I haven't taken a course on calculus so far so I don't know what to do. The integral may be wrong. Please tell me which part of it is wrong.
$$
q∫_{+a}^{-a}\lim_{c \to g}\frac 1{(b^2+c^2)^{3⁄2}} dc
$$
All letters in the problem are variables.
Since the question can't be understood. Here is the figure of the main context:

b: radius of the cylinder
+a and -a: the x component of the cylinder's corners' positon vectors
Actually this is an electrostatic problem. I am trying to find the electric field due to a tube applied on a point. Since the tube is infinite number of rings, I first found the electric field of a ring which is
$E=kcq/(c^2+b^2)^{(3⁄2)}$
where k=8,98... is Coulomb's constant and q is the charge of an infinitesmal ring. And g is the distance between the point and the +a.
Hope this helps.
 A: You have a ring, with a center in $(0,0,0)$. And on the point on the axis it has induce the electric field:
$$
E = \frac{q z}{4\pi\varepsilon_0 (r^2 + z^2)^{3/2}}
$$
where, $r$ is a ring radius and $z$ is a distance from origin, so, point have coordinates $(0,0,z)$.
Then, if you move the ring by some factor $\delta$ it become and $\delta$ would be between $-a$ and $a$:
$$
E = \frac{q (z + \delta)}{4\pi\varepsilon_0 (r^2 + (z+\delta)^2)^{3/2}}
$$
Now, it's only needed to integrate this by $\delta$ from $-a$ to $a$:
$$
E = \int_{-a}^{a}\frac{q (z + \delta)}{4\pi\varepsilon_0 (r^2 + (z+\delta)^2)^{3/2}}\mathrm{d}\delta = \frac{q}{4\pi\varepsilon_0}\int_{-a}^{a}\frac{z+\delta}{(r^2 + (z+\delta)^2)^{3/2}}
$$
Integrating this:
$$
E = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{\sqrt{r^2 + (z-a)^2}} - \frac{1}{\sqrt{r^2 + (z+a)^2}}\right)
$$
A: Note that, for every $(b,g)\ne(0,0)$,
$$
\lim_{c \to g}\frac 1{(b^2+c^2)^{3⁄2}}=\frac 1{(b^2+g^2)^{3⁄2}}
$$ 
hence
$$
I=q∫_{+a}^{-a}\lim_{c \to g}\frac 1{(b^2+c^2)^{3⁄2}} dc,$$ is $$I=q∫_{+a}^{-a}\frac 1{(b^2+g^2)^{3⁄2}} dc=\frac q{(b^2+g^2)^{3⁄2}}∫_{+a}^{-a} dc=\frac{-2aq}{(b^2+g^2)^{3⁄2}}.
$$
