Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{e^{2 x}-5 e^{w}+4}{e^{w}-1} d w$$

Short answer

Question: Find the indefinite integral of the following function: $$\int \frac{e^{2 x}-5 e^{w}+4}{e^{w}-1} dw$$ Solution: The indefinite integral of the given function is: $$\int \frac{e^{2 x}-5 e^{w}+4}{e^{w}-1} dw = e^w + 5w + C$$ where \(C\) is the constant of integration.

Step-by-step solution

Simplify the integrand

First, let's rewrite the given integral with a clearer denominator: $$\int \frac{e^{2 x}-5 e^{w}+4}{e^w - 1} dw$$ Now, we perform synthetic division on the numerator with respect to the denominator, \(e^w - 1\): $$\frac{e^{2x}-5 e^{w}+4}{e^w - 1} = e^w + 5$$ This step simplifies our integrand, allowing us to rewrite the integral as: $$\int \left(e^w + 5\right) dw$$

Integrate the simplified integrand

Now we can easily integrate each term separately: $$\int \left(e^w + 5\right) dw = \int e^w dw + \int 5 dw$$ The indefinite integral of \(e^w\) is simply \(e^w\) and the indefinite integral of \(5\) is \(5w\). Therefore, the resulting integral is: $$e^w + 5w + C$$ Where \(C\) is the constant of integration.

Check the result by differentiation

To ensure our solution is correct, we take the derivative of the integrated function with respect to \(w\): $$\frac{d}{dw}\left(e^w + 5w + C\right)$$ The derivative of \(e^w\) is \(e^w\), the derivative of \(5w\) is \(5\), and the derivative of the constant \(C\) is \(0\). Thus, the expression becomes: $$e^w + 5$$ Since the derived expression matches our simplified integrand from Step 1, our solution for the indefinite integral is correct. Therefore, the final indefinite integral is: $$\int \frac{e^{2 x}-5 e^{w}+4}{e^{w}-1} dw = e^w + 5w + C$$

Key concepts

Synthetic Division

Synthetic division is a simplified form of polynomial long division. It is useful when we're working with polynomials and want to divide by a linear factor. In other words, it's a method of breaking down a complex fraction into simpler parts. In the context of indefinite integrals, simplifying the integrand using synthetic division can make subsequent integration steps much more manageable.

To apply synthetic division to a given integrand, follow these steps:

• Identify the numerator and the denominator of your function. In this case, our numerator is a combination of exponential functions like \(e^{2x} - 5e^{w} + 4\), and the denominator is \(e^w - 1\).
• Simplify the fraction using synthetic division, dividing the polynomials accordingly. This means adjusting the post-division terms so that the numerator matches the denominator.

By performing this division, what initially looks complex becomes straightforward: \(e^w + 5\). Now, the integration process can proceed with these simplified terms.

Integration by Parts

Integration by parts is a technique derived from the product rule for derivatives. It helps solve integrals where the standard methods like substitution aren't effective. We use it when we encounter a product of two functions, say \(u\) and \(v\), that makes a direct integration difficult. The formula for integration by parts is given by:

\[ \int u \, dv = uv - \int v \, du \]

This technique is not directly applied in the given problem, but it is invaluable for other complex integrals that involve a product of functions. Here's a quick guide on using integration by parts:

• Choose which part of the integrand to set as \(u\). Usually, a function that becomes simpler when differentiated. Logarithms and inverse trigonometric functions are common choices.
• Select the remaining part as \(dv\). It should be easily integrable.
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\).
• Plug these into the integration by parts formula to solve the integral.

Integration by parts often simplifies an integral that seems impossible to integrate directly, making it an essential tool in calculus.

Constant of Integration

When working with indefinite integrals, one crucial element to remember is the constant of integration, denoted as \(C\). This constant arises because integration is the reverse process of differentiation, and there's no unique way to determine the original constant value from the function's derivative.

Here’s why the constant is important:

• Any function \(f(x)\) has infinitely many antiderivatives, each differing by a constant value. Hence, \(\int f(x) \, dx = F(x) + C\), where \(F(x)\) is the antiderivative of \(f(x)\).
• The constant \(C\) accounts for all possible vertical shifts of the antiderivative, ensuring all possible values are represented.
• When solving differential equations or other applications, determining \(C\) may require initial conditions or boundary values.

In practice, always remember to include \(C\) in your indefinite integral answers to represent the full family of possible solutions. This is essential for maintaining the mathematical correctness of your result.