Question:
Verify the following
$\int \frac{2 x-1}{2 x+3} d x=x-\log \left|(2 x+3)^{2}\right|+C$
Solution:
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