Evaluate the integral: integral x2 e^-x\,dx

Choose $u$ and $dv$:
Let $u = x^2$
Then $du = 2x\,dx$

Let $dv = e^{-x}\,dx$
Then $v = \int e^{-x}\,dx = -e^{-x}$

Apply the integration by parts formula:

$$\int x^2 e^{-x}\,dx = x^2(-e^{-x}) - \int (-e^{-x})(2x)\,dx$$ $$\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2 \int x e^{-x}\,dx$$

Choose new $u$ and $dv$:
Let $u_1 = x$
Then $du_1 = dx$

Let $dv_1 = e^{-x}\,dx$
Then $v_1 = \int e^{-x}\,dx = -e^{-x}$

Apply the integration by parts formula to $\int x e^{-x}\,dx$:

$$\int x e^{-x}\,dx = x(-e^{-x}) - \int (-e^{-x})\,dx$$ $$\int x e^{-x}\,dx = -x e^{-x} + \int e^{-x}\,dx$$ $$\int x e^{-x}\,dx = -x e^{-x} - e^{-x}$$

$$\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2 (-x e^{-x} - e^{-x}) + C$$ $$\int x^2 e^{-x}\,dx = -x^2 e^{-x} - 2x e^{-x} - 2 e^{-x} + C$$

We can factor out $-e^{-x}$:

$$\int x^2 e^{-x}\,dx = -e^{-x}(x^2 + 2x + 2) + C$$

The final answer is $\boxed{-e^{-x}(x^2 + 2x + 2) + C}$.

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