Integral of xe^(-x/2) dx

1. Use integration by parts:

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

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

   Then $\operatorname{du}{\left(x \right)} = 1$.

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

   1. There are multiple ways to do this integral.

      Method #1

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

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

         $$\int 4 e^{u}\, du$$

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

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

            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:

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

      Method #2

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

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

         $$\int 4\, du$$

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

            $$\int \left(-2\right)\, du = - 2 \int 1\, du$$

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            So, the result is: $- 2 u$

         Now substitute $u$ back in:

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

   Now evaluate the sub-integral.

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

   $$\int \left(- 2 e^{- \frac{x}{2}}\right)\, dx = - 2 \int e^{- \frac{x}{2}}\, dx$$

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

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

      $$\int 4 e^{u}\, du$$

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

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

         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: $4 e^{- \frac{x}{2}}$

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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