Integral of e^(0,5x-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$e^{\frac{x}{2} - 1} \cdot 1 = \frac{e^{\frac{x}{2}}}{e}$$

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

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

      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:

               $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

               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:

               $$\int 2\, du = 2 \int 1\, du$$

               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:

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

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

   Method #2

   1. Rewrite the integrand:

      $$e^{\frac{x}{2} - 1} \cdot 1 = \frac{e^{\frac{x}{2}}}{e}$$

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

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

      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:

            $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

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

2. Now simplify:

   $$2 e^{\frac{x}{2} - 1}$$

3. Add the constant of integration:

   $$2 e^{\frac{x}{2} - 1}+ \mathrm{constant}$$

The answer is:

$$2 e^{\frac{x}{2} - 1}+ \mathrm{constant}$$