Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{e^{2 t}-1}{e^{t}-1} d t$$

Short answer

Question: Find the indefinite integral of the function $$\frac{e^{2 t}-1}{e^{t}-1}$$ with respect to \(t\), and verify your answer by differentiation. Answer: The indefinite integral of the function $$\frac{e^{2 t}-1}{e^{t}-1}$$ with respect to \(t\) is $$e^t + C$$, where \(C\) is the constant of integration. This can be verified by differentiation, as differentiating $$e^t + C$$ with respect to \(t\) yields the simplified function $$e^t$$.

Step-by-step solution

Simplify the integrand if possible

In this case, we can factor out \(e^t\) from the numerator to simplify the expression: $$\frac{e^{2t}-1}{e^t-1} = \frac{e^t(e^t-1)}{e^t-1}$$ We can now cancel out the common factor \((e^t-1)\): $$\frac{e^{2t}-1}{e^t-1} = e^t$$ Now we need to find the integral of the simplified function: $$\int\frac{e^{2t}-1}{e^t-1} dt = \int e^t dt$$

Apply the integral

The integral of the exponential function \(e^t\) is simple and straightforward: $$\int e^t dt = e^t + C$$ where \(C\) is the constant of integration. Now let's check our work by differentiating the result.

Differentiate the result to check

Differentiating our result with respect to \(t\): $$\frac{d}{dt}(e^t + C) = e^t$$ Since the result of our differentiation is the same as our simplified function, our integration is correct: $$\int \frac{e^{2 t}-1}{e^{t}-1} d t = e^t + C$$

Key concepts

Simplifying Integrands

When approaching an indefinite integral, it is often helpful to simplify the integrand as much as possible before attempting to find the antiderivative. Simplifying the integrand involves transforming it into an easier form that will make the integration process more straightforward. In the given exercise, we have:

• The original integrand is \[ \frac{e^{2t} - 1}{e^t - 1} \]
• Recognizing that the numerator, \(e^{2t} - 1\), can be rewritten by factoring an \(e^t\) out, gives us:\[ \frac{e^t(e^t - 1)}{e^t - 1} \]
• This allows us to cancel the \(e^t - 1\) term in the numerator and denominator, resulting in simply \(e^t\).

This simplification reduces the complexity of the integral, allowing us to easily integrate the simplified expression \(e^t\). Simplifying integrands is a powerful technique in calculus that reduces cumbersome algebraic manipulation and focuses on straightforward integration.

Exponential Functions

Exponential functions, particularly with base \(e\), are crucial components in calculus. The function \(e^t\) is a fundamental exponential function, where \(e\) is Euler's number, approximately 2.718.

• One of the key properties of the exponential function \(e^t\) is its unique derivative and integral behavior: the derivative or integral of \(e^t\) is itself, \(e^t\).
• This property simplifies many calculus operations, making \(e^t\) a common and convenient choice for modeling continuous growth or decay processes.
• In the context of the integral \(\int e^t \, dt\), the solution is straightforward:\[ e^t + C \]where \(C\) is the constant of integration, representing the family of antiderivatives.

Thus, understanding the unique properties of exponential functions aids in seamlessly performing integration and differentiation, especially when dealing with functions involving \(e\).

Differentiation Check

Once an indefinite integral has been solved, verifying the correctness of your solution is essential. A common method is through differentiation, essentially reversing the integration process to ensure the original function is obtained.

• For the integral \[ \int e^t \, dt = e^t + C \]averifying by differentiation means taking the derivative of the result, \(e^t + C\).
• Using derivative rules, the derivative of \(e^t + C\) results in:\[ \frac{d}{dt}(e^t + C) = e^t \]
• Since the differentiation result \(e^t\) matches the simplified integrand, this confirms the original integration was performed correctly.

Differentiation checks not only strengthen understanding but also offer reassurance that each step in solving indefinite integrals adheres to the core principles of calculus.