Q. 51 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \frac{2-e^x}{e^x}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{2-e^x}{e^x}\mathrm{d}x=-2e^{-x}-x+C$.

Step 1. Given Information

Solving the given integrals.

$\int \frac{2-e^x}{e^x}\mathrm{d}x$

Step 2. Firstly simplifying the integral.

$$\begin{gathered} \int \frac{2-e^x}{e^x}\mathrm{d}x=\int \left(\frac{2}{e^x}-\frac{e^x}{e^x}\right)\mathrm{d}x \\ \int \frac{2-e^x}{e^x}\mathrm{d}x=\int \left(2e^{-x}-1\right)\mathrm{d}x \\ \int \frac{2-e^x}{e^x}\mathrm{d}x=2\int e^{-x}\mathrm{d}x-\int 1\mathrm{d}x \\ \int \frac{2-e^x}{e^x}\mathrm{d}x=-2e^{-x}-x+C \end{gathered}$$