Integral Ln 2X+1 Aae674

1. **State the problem:** Evaluate the integral $$\int \ln(2x + 1) \, dx$$. 2. **Formula and method:** Use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = \ln(2x + 1) \quad \Rightarrow \quad du = \frac{2}{2x + 1} dx$$ $$dv = dx \quad \Rightarrow \quad v = x$$ 3. **Apply integration by parts:** $$\int \ln(2x + 1) \, dx = x \ln(2x + 1) - \int x \cdot \frac{2}{2x + 1} dx$$ 4. **Simplify the integral:** $$\int x \cdot \frac{2}{2x + 1} dx = 2 \int \frac{x}{2x + 1} dx$$ 5. **Rewrite the integrand:** Express $$\frac{x}{2x + 1}$$ as: $$\frac{x}{2x + 1} = \frac{1}{2} - \frac{1}{2(2x + 1)}$$ 6. **Split the integral:** $$2 \int \frac{x}{2x + 1} dx = 2 \int \left( \frac{1}{2} - \frac{1}{2(2x + 1)} \right) dx = 2 \left( \int \frac{1}{2} dx - \int \frac{1}{2(2x + 1)} dx \right)$$ 7. **Simplify constants:** $$= 2 \left( \frac{1}{2} \int dx - \frac{1}{2} \int \frac{1}{2x + 1} dx \right) = \int dx - \int \frac{1}{2x + 1} dx$$ 8. **Integrate each term:** $$\int dx = x$$ $$\int \frac{1}{2x + 1} dx = \frac{1}{2} \ln|2x + 1|$$ 9. **Combine results:** $$2 \int \frac{x}{2x + 1} dx = x - \frac{1}{2} \ln|2x + 1| + C$$ 10. **Final answer:** $$\int \ln(2x + 1) \, dx = x \ln(2x + 1) - \left( x - \frac{1}{2} \ln|2x + 1| \right) + C = x \ln(2x + 1) - x + \frac{1}{2} \ln|2x + 1| + C$$