Answer
$-\dfrac{1}{4(4x-3)}+C$
Work Step by Step
We are given that $I=\int \dfrac{\ dx}{(4x-3)^2}$
In order to solve the above integral, we will use the following formula such as:
$\int (ax+b)^n \ dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+C$
Now, we have $\int \dfrac{\ dx}{(4x-3)^2} \ dx=\int (4x-3)^{-2} \ dx$
or, $=\int \dfrac{(4x-3)^{-2+1}}{4(-2+1)}+C $
or, $=-\dfrac{1}{4(4x-3)}+C$
