Question

Find the indefinite integral. $$ \int \frac{e^{2 x}+2 e^{x}+1}{e^{x}} d x $$

Short answer

The indefinite integral \( \int \frac{e^{2x} + 2e^{x} + 1}{e^{x}} dx \) equals \( e^{x} + 2x - e^{-x} + C \), where 'C' represents the constant of integration.

Step-by-step solution

Split the Integral

Notice that the integral can be separated into simpler fractions to make it easier to solve. So break down the fraction: \[ \int \frac{e^{2 x}+2 e^{x} + 1}{e^{x}} dx = \int e^{x} dx + \int 2 dx + \int \frac{1}{e^{x}} dx \]

Compute The First Integral

The integration of \( e^{x} \) is simple, it will just stay as \( e^{x} \). So, \[ \int e^{x} dx = e^{x} \]

Compute The Second Integral

The integral \[ \int 2 dx \] is also simple. The integral of a constant 'a' is just 'a*x,' and in this case 'a' happens to be 2. So, \[ \int 2 dx = 2x \]

Compute The Third Integral

The integration of an exponential function with a negative exponent can be computed using the formula for the integral of \( e^{ax} \), which is \( \frac{1}{a} e^{ax} \). In this case, 'a' is -1, so \[ \int \frac{1}{e^{x}} dx = - e^{-x} \]

Add the Constants of Integration

Each time you perform an indefinite integral, you add a constant of integration. However, instead of adding separate constants for each integral computed, you can add a single general constant at the end. This is because the sum of the constants of integration would be a constant itself.

Combine the Results

Combine all of the computations from the previous steps into a single equation to find the result of the original integral. Therefore, the final result is \[ e^{x} + 2x - e^{-x} + C \] where 'C' is the constant of integration.