If $\int\frac{2x-5}{(2x-3)^3}e^{2x}dx=\frac{λ e^{2x}}{(2x-3)^2}+C$, where c is an arbitrary constant then the value of $λ$ is 

$\frac{1}{2}$

The correct answer is Option (1) → $\frac{1}{2}$

$\int \frac{2x-5}{(2x-3)^3}e^{2x}dx=\frac{\lambda e^{2x}}{(2x-3)^2}+C.$

$\text{Differentiate RHS:}$

$\frac{d}{dx}\left(\frac{\lambda e^{2x}}{(2x-3)^2}\right) =\lambda\left[\frac{2e^{2x}}{(2x-3)^2}+e^{2x}\frac{d}{dx}(2x-3)^{-2}\right].$

$=\lambda\left[\frac{2e^{2x}}{(2x-3)^2}-\frac{4e^{2x}}{(2x-3)^3}\right].$

$=\lambda e^{2x}\frac{4x-10}{(2x-3)^3}.$

$=\lambda e^{2x}\frac{2(2x-5)}{(2x-3)^3}.$

$=2\lambda\frac{2x-5}{(2x-3)^3}e^{2x}.$

$\text{Equating with integrand:}$

$2\lambda=1.$

$\lambda=\frac{1}{2}.$

$\lambda=\frac{1}{2}.$