Verify the following

L.H.S. $=\int \frac{2 x-1}{2 x+3} d x$

$=\int\left(1-\frac{4}{2 x+3}\right) d x$     [Dividing the numerator by the denominator]

$=\int 1 \cdot d x-4 \int \frac{1}{2 x+3} d x=\int 1 \cdot d x-\frac{4}{2} \int \frac{1}{x+\frac{3}{2}} d x$

$=\int 1 \cdot d x-2 \int \frac{1}{x+\frac{3}{2}} d x=x-2 \log \left|x+\frac{3}{2}\right|+C$

$=x-2 \log \left|\frac{2 x+3}{2}\right|+C=x-\log \left|\left(\frac{2 x+3}{2}\right)^{2}\right|+C$

$\left[\because n \log m=\log m^{n}\right]$

$=x-\log \left|(2 x+3)^{2}\right|-\log 2^{2}+C$

$=x-\log \left|(2 x+3)^{2}\right|+C_{1} \Rightarrow$ R.H.S. $\quad\left[\right.$ where $\left.C_{1}=C-\log 2^{2}\right]$

L.H.S. = R.H.S.

- Hence Proved