Integral of exp((4*x)/3)-(2/exp(-x)) dx

1. Integrate term-by-term:

   1. Let $u = \frac{4 x}{3}$.

      Then let $du = \frac{4 dx}{3}$ and substitute $\frac{3 du}{4}$:

      $$\int \frac{3 e^{u}}{4}\, du$$

      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: $\frac{3 e^{u}}{4}$

      Now substitute $u$ back in:

      $$\frac{3 e^{\frac{4 x}{3}}}{4}$$

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

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

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

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

         $$\int \left(- \frac{1}{u^{2}}\right)\, du$$

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

            $$\int \frac{1}{u^{2}}\, du = - \int \frac{1}{u^{2}}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

            So, the result is: $\frac{1}{u}$

         Now substitute $u$ back in:

         $$e^{x}$$

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

   The result is: $- 2 e^{x} + \frac{3 e^{\frac{4 x}{3}}}{4}$

2. Now simplify:

   $$\frac{3 e^{\frac{4 x}{3}}}{4} - 2 e^{x}$$

3. Add the constant of integration:

   $$\frac{3 e^{\frac{4 x}{3}}}{4} - 2 e^{x}+ \mathrm{constant}$$

The answer is: