1. We are asked to find the integral $$\int (r^2 + r + 1)e^r \, dr$$. 2. To solve this, we use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$. 3. Let’s choose parts: - Let $$u = r^2 + r + 1$$, so $$du = (2r + 1) dr$$. - Let $$dv = e^r dr$$, so $$v = e^r$$. 4. Applying integration by parts: $$\int (r^2 + r + 1)e^r dr = (r^2 + r + 1)e^r - \int (2r + 1)e^r dr$$. 5. Now, we need to evaluate $$\int (2r + 1)e^r dr$$. Use integration by parts again: - Let $$u = 2r + 1$$, so $$du = 2 dr$$. - Let $$dv = e^r dr$$, so $$v = e^r$$. 6. Applying integration by parts again: $$\int (2r + 1)e^r dr = (2r + 1)e^r - \int 2 e^r dr = (2r + 1)e^r - 2 \int e^r dr$$. 7. Since $$\int e^r dr = e^r$$, we have: $$\int (2r + 1)e^r dr = (2r + 1)e^r - 2 e^r = (2r + 1 - 2)e^r = (2r - 1)e^r$$. 8. Substitute back: $$\int (r^2 + r + 1)e^r dr = (r^2 + r + 1)e^r - (2r - 1)e^r + C = (r^2 + r + 1 - 2r + 1)e^r + C = (r^2 - r + 2)e^r + C$$. 9. Final answer: $$\boxed{\int (r^2 + r + 1)e^r dr = (r^2 - r + 2)e^r + C}$$. This integral was solved using integration by parts twice, carefully choosing $$u$$ and $$dv$$ each time, and simplifying the resulting expressions.