Two identical thin rods each of length $2l$ have equal mass m, that is distributed uniformly along their lengths. The rods lie along X-axis with their centres separated by a distance  $d>2l$

The magnitude of the force exerted by the left rod on the right one is  $\frac{G{m}^2}{4l^2}\ln \left[\frac{d^2}{d^2-\beta l^2}\right]$. Find the value of  $\beta$.

Consider only left rod (in the absence of right rod). Consider a hypothetical point P as shown 

Gravitational field due to element, at P is  $d{E}_g=\frac{Gdm}{X^2}=\frac{G^{\frac{m}{2l}}}{X^2}dx$
At P,  $E_g=\frac{Gm}{2l}{\int }_{d^1}^{d^1+2l}\frac{1}{X^2}dx=-\frac{Gm}{2l}{\left[\frac{1}{X}\right]}_{d^1}^{d^1+2l}$
 $\therefore E_g=\frac{Gm}{2l}[\frac{1}{d^1}-\frac{1}{d^1+2l}]=\frac{Gm}{2l}\frac{d^1+2l-d^1}{d^1(d^1+2l)}$
Or $E_g=\frac{Gm}{d^1(d^1+2l)}$ ……(i) Now, let us consider two rods
 
Consider an element of length dx on right side rod. This element is at a distance x from right end of left rod. Field due to left rod at the location of this element is  $\frac{Gm}{x(x+2l)}$ by using Eq.(i) Force on this element is = $dF=E_{g}dm\frac{Gm}{x(x+2l)}.\frac{m}{2l}dx$  or  $dF=\frac{G{m}^2}{2l}.\frac{1}{x(x+2l)}dx$