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

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

Differentiating the right side:

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

Let $u = \lambda e^{2x}$ and $v = (2x-3)^2$

Then $u' = 2\lambda e^{2x}$ and $v' = 4(2x-3)$

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Applying the quotient rule:

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

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

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

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

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Simplifying the numerator:

$2(2x-3) - 4 = 4x - 6 - 4$

$= 4x - 10$

$= 2(2x - 5)$

Therefore:

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

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The derivative must equal the original integrand:

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

Comparing coefficients:

$2\lambda = 1$

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