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(D) We use the integration by parts formula: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$. Let $u = (x+1)^2$ and $v = e^x$. Then $u' = 2(x

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$(x^2+1) e^x+C$

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(D) We use the integration by parts formula: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$. Let $u = (x+1)^2$ and $v = e^x$. Then $u' = 2(x+1)$ and $\int v d x = e^x$. Applying the formula: $\int(x+1)^2 e^x d x = (x+1)^2 e^x - \int 2(x+1) e^x d x$. Now, apply integration by parts again to $\int (x+1) e^x d x$: $\int (x+1) e^x d x = (x+1) e^x - \int 1 \cdot e^x d x = (x+1) e^x - e^x$. Substituting this back: $= (x+1)^2 e^x - 2[(x+1) e^x - e^x] + C$ $= (x^2+2x+1) e^x - 2x e^x - 2 e^x + 2 e^x + C$ $= (x^2+2x+1-2x) e^x + C$ $= (x^2+1) e^x + C$.

$\int(x+1)^2 e^x d x$ is equal to

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