Integral of (x*x)*exp(-x/2) dx

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{- \frac{x}{2}}$.

   Then $\operatorname{du}{\left(x \right)} = 2 x$.

   To find $v{\left(x \right)}$:

   1. Let $u = - \frac{x}{2}$.

      Then let $du = - \frac{dx}{2}$ and substitute $- 2 du$:

      $$\int \left(- 2 e^{u}\right)\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\text{False}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $- 2 e^{u}$

      Now substitute $u$ back in:

      $$- 2 e^{- \frac{x}{2}}$$

   Now evaluate the sub-integral.

2. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = - 4 x$ and let $\operatorname{dv}{\left(x \right)} = e^{- \frac{x}{2}}$.

   Then $\operatorname{du}{\left(x \right)} = -4$.

   To find $v{\left(x \right)}$:

   1. Let $u = - \frac{x}{2}$.

      Then let $du = - \frac{dx}{2}$ and substitute $- 2 du$:

      $$\int \left(- 2 e^{u}\right)\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\text{False}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $- 2 e^{u}$

      Now substitute $u$ back in:

      $$- 2 e^{- \frac{x}{2}}$$

   Now evaluate the sub-integral.

3. The integral of a constant times a function is the constant times the integral of the function:

   $$\int 8 e^{- \frac{x}{2}}\, dx = 8 \int e^{- \frac{x}{2}}\, dx$$

   1. Let $u = - \frac{x}{2}$.

      Then let $du = - \frac{dx}{2}$ and substitute $- 2 du$:

      $$\int \left(- 2 e^{u}\right)\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\text{False}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $- 2 e^{u}$

      Now substitute $u$ back in:

      $$- 2 e^{- \frac{x}{2}}$$

   So, the result is: $- 16 e^{- \frac{x}{2}}$

4. Now simplify:

   $$- \left(2 x^{2} + 8 x + 16\right) e^{- \frac{x}{2}}$$

5. Add the constant of integration:

   $$- \left(2 x^{2} + 8 x + 16\right) e^{- \frac{x}{2}}+ \mathrm{constant}$$

The answer is:

$$- \left(2 x^{2} + 8 x + 16\right) e^{- \frac{x}{2}}+ \mathrm{constant}$$