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(D) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = (x+1)^2$ and $dv = e^x \, dx$. Then $du = 2(x+1) \, dx$ an

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

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(D) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = (x+1)^2$ and $dv = e^x \, dx$. Then $du = 2(x+1) \, dx$ and $v = e^x$. $\int (x+1)^2 e^x \, dx = (x+1)^2 e^x - \int 2(x+1) e^x \, dx$. Now,apply integration by parts again for $\int (x+1) e^x \, dx$: Let $u = (x+1)$ and $dv = e^x \, dx$. Then $du = dx$ and $v = e^x$. $\int (x+1) e^x \, dx = (x+1) e^x - \int e^x \, dx = (x+1) e^x - e^x = x e^x$. Substituting this back: $\int (x+1)^2 e^x \, dx = (x+1)^2 e^x - 2(x e^x) + C$ $= (x^2 + 2x + 1) e^x - 2x e^x + C$ $= (x^2 + 1) e^x + C$.

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

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