Browse Questions

# Verify the following$\int\frac{2x-1}{2x+1}dx=x-log \mid (2x+3)^2\mid+C$

Toolbox:
• If the degree of the numerator of rational expression is equal to or greater than the degree of the denominator, then it is said to be improper
• $\int \frac{dx}{(x+a)}=log |x+a|+c$
Let $I=\large\int \frac{2x-1}{2x+3}dx$
Since the numerator and the denominator are of the same degree, it is an improper rational expression. so let us divide it to make it a proper rational expression.
Therefore $\large\frac{2x-1}{2x+3}=1-\frac{4}{2x-3}$
Hence $I=\int dx-\int \large\frac{4}{2x+3}dx$
on integrating we get
$x-4 \times \frac{1}{2} log (2x+3)+c$
$=x-2 log (2x+3)+c$
$=x-log (2x+3)^2+c$
=R.H.S
Hence L.H.S=R.H.S is proved