Integral of 1/2xe^(-x/2) dx

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

   $$\int \frac{x e^{\frac{\left(-1\right) x}{2}}}{2}\, dx = \frac{\int x e^{\frac{\left(-1\right) x}{2}}\, dx}{2}$$

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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