Find: `int (2x-5)/(2x-3)^3 e^(2x) dx.`

Answer

To solve the integral \( \int \frac{2x-5}{(2x-3)^3} e^{2x} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = 2x - 3 \). Then, we differentiate to find \( dx \):
\[ dt = 2 \, dx \implies dx = \frac{1}{2} dt \]
Also, we can express \( 2x \) in terms of \( t \):
\[ 2x = t + 3 \implies 2x - 5 = t + 3 - 5 = t - 2 \]

### Step 2: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ \int \frac{2x-5}{(2x-3)^3} e^{2x} \, dx = \int \frac{t-2}{t^3} e^{t+3} \cdot \frac{1}{2} dt \]
This simplifies to:
\[ \frac{1}{2} \int \frac{t-2}{t^3} e^{t+3} \, dt \]

### Step 3: Simplify the Integral
We can separate the integral:
\[ \frac{1}{2} \int \left( \frac{t}{t^3} - \frac{2}{t^3} \right) e^{t+3} \, dt = \frac{1}{2} \int \left( \frac{1}{t^2} - \frac{2}{t^3} \right) e^{t+3} \, dt \]
This can be rewritten as:
\[ \frac{1}{2} e^3 \int \left( \frac{1}{t^2} - \frac{2}{t^3} \right) e^t \, dt \]

### Step 4: Identify \( f(t) \) and \( f'(t) \)
Let \( f(t) = \frac{1}{t^2} \). Then, we find \( f'(t) \):
\[ f'(t) = -\frac{2}{t^3} \]
Thus, we can express our integral as:
\[ \frac{1}{2} e^3 \int e^t \left( f(t) + f'(t) \right) dt \]

### Step 5: Integrate Using the Formula
Using the integration formula \( \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C \), we get:
\[ \frac{1}{2} e^3 \left( e^t f(t) + C \right) = \frac{1}{2} e^3 \left( e^t \cdot \frac{1}{t^2} + C \right) \]

### Step 6: Substitute Back
Substituting back \( t = 2x - 3 \):
\[ = \frac{1}{2} e^3 \left( \frac{e^{2x-3}}{(2x-3)^2} + C \right) \]
This simplifies to:
\[ = \frac{e^{2x}}{2(2x-3)^2} + C' \]
where \( C' = \frac{e^3}{2}C \).

### Final Answer
Thus, the final result is:
\[ \int \frac{2x-5}{(2x-3)^3} e^{2x} \, dx = \frac{e^{2x}}{2(2x-3)^2} + C \]