1. The problem involves finding the integral using integration by parts where $u=\ln(x+2)$ and $dv=dx$. 2. Recall the integration by parts formula: $$\int u\,dv = uv - \int v\,du$$. 3. Given $u=\ln(x+2)$, differentiate to find $du$: $$du = \frac{1}{x+2} dx$$. 4. Given $dv=dx$, integrate to find $v$: $$v = x$$. 5. Substitute into the formula: $$\int \ln(x+2) dx = x \ln(x+2) - \int x \cdot \frac{1}{x+2} dx$$. 6. Simplify the integral: $$\int \frac{x}{x+2} dx$$. 7. Rewrite the integrand: $$\frac{x}{x+2} = \frac{x+2-2}{x+2} = 1 - \frac{2}{x+2}$$. 8. So the integral becomes: $$\int 1 dx - 2 \int \frac{1}{x+2} dx = x - 2 \ln|x+2| + C$$. 9. Substitute back to get the final answer: $$x \ln(x+2) - (x - 2 \ln|x+2|) + C = x \ln(x+2) - x + 2 \ln|x+2| + C$$. 10. This is the integral of $\ln(x+2)$ with respect to $x$ using integration by parts.