Finding the integral $\int_{0}^{\infty }\frac{1}{(4x-3)(4x-1)}\,dx$. Which method that will be effective for solving this integral?

The integral does not converge because of the singularities.
To compute the integral between $a$ and $\infty$ for $a>\frac 34$, you can use the following method:
\begin{align}
\int_{a}^{\infty }\frac{1}{(4x-3)(4x-1)} dx
&= \lim_{N\to\infty} \int_{a}^N \frac{1}{(4x-3)(4x-1)} dx
\end{align}
Now use the fact that $$
\frac{1}{(4x-3)(4x-1)} =
\frac 12 \left(
\frac 1{4x-3} - \frac 1{4x-1}
\right)
$$
so that
\begin{align}
\int_{a}^N \frac{dx}{(4x-3)(4x-1)} 
&= \frac 12 \left(
\int_{a}^N \frac {dx}{4x-3} dx - \int_a^N \frac {dx}{4x-1} dx
\right) 
\\&= \frac 18\left(
\log \frac{|4N-3|}{|4a-3|} - \log \frac{|4N-1|}{|4a-1|}
\right)
\\&= \frac 18\left(
\log \frac{4N-3}{4N-1} - \log \frac{4a-3}{4a-1}\right)
 \to -\frac18 \log \frac{4a-3}{4a-1}
\end{align}
