Step-by-Step Solution
Key Concept: General
$$\int \ln(2x+3)^{(2x+3)} \, dx = \int \underbrace{(2x+3)}_{II} \underbrace{\ln(2x+3)}_{I} \, dx$$ $$= \ln(2x+3) \int (2x+3) \, dx - \int \frac{1}{(2x+3)} \left( \int (2x+3) \, dx \right) dx$$ $$= \ln(2x+3) \frac{(2x+3)^2}{4} - \int \frac{1}{2x+3} \times \frac{(2x+3)^2}{2} \, dx$$ $$= \ln(2x+3) \frac{(2x+3)^2}{4} - \frac{(2x+3)^2}{4} + C$$
Correct Answer: $\frac{(2x+3)^2}{4} \ln(2x+3) - \frac{(2x+3)^2}{4} + C$