Why do you need to use the chain rule in differentiation of ln? I understand application of chain rule in the differentiation of a random function $(x^2+3)^3$. 
But, why do you need to use chain-rule when differentiating something like $\ln(2x-1)$; why won't it just be $\displaystyle\frac 1{2x-1}$? Please help.

Let us first call
$$y = \ln(2x + 1)$$
Physicists and engineers use the simpler Leibniz calculus to calculate the differential quotient $\frac{dy}{dx}$ instead of using those pesky Newtonian fluxion dots ($\dot{y}$) or French apostrophes ($y'$) .. :-)
We substitute 
$$u = 2x + 1$$
and get
$$y = \ln(u)$$
and 
$$\frac{dy}{dx} = \frac{d\ln(u)}{dx} = \frac{d\ln(u)}{du}\frac{du}{dx}$$.
With
$$\frac{d\ln(u)}{du} = \frac{1}{u}$$
and
$$\frac{du}{dx} = \frac{d(2x + 1)}{dx} = 2$$
we get
$$\frac{dy}{dx} = \frac{1}{2x + 1} \cdot 2$$
Here we used the chain rule in its guise as substition rule (or coordinate transformation $u \to x$), canceling differentials in a fraction. The important bit is the factor 2 when transforming from $u$ to $x$ coordinates.
The substituion technique is even more helpful when solving integrals.
$$y = \int dy + C = \int \frac{dy}{dx} dx + C = \int \frac{2}{2x+1} dx + C = \int \frac{du}{u} + C = \ln u = \ln(2x + 1)$$
for some constant $C$. 
