Problem 37 - Applied Calculus

Question

Find each indefinite integral. \(\int \frac{x^{2}-1}{x+1} d x\)

Step-by-Step Solution

Answer

The integral \( \int \frac{x^2 - 1}{x+1} \, dx \) is \( \frac{x^2}{2} - x + C \).

Step 1: Simplify the integrand using long division

To simplify the integrand \( \frac{x^2 - 1}{x + 1} \), perform polynomial long division. Divide \( x^2 - 1 \) by \( x + 1 \). First, divide the leading term \( x^2 \) by \( x \) to get \( x \). Multiply \( x \) by \( x + 1 \) which gives \( x^2 + x \). Subtract this from \( x^2 - 1 \) to get \(-x - 1 \). Divide \(-x \) by \( x \) to get \(-1 \). Multiply \(-1\) by \( x + 1 \) to get \(-x - 1 \). Subtract this result to get \( 0 \). Therefore, the integrand simplifies to \( x - 1 \).

Step 2: Integrate the simplified expression

Now we integrate the simplified expression \( x - 1 \). The indefinite integral becomes \( \int (x - 1) \, dx \).

Step 3: Apply the power rule of integration

Using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), integrate \( x \) and \( -1 \). The integral of \( x \) is \( \frac{x^2}{2} \). The integral of \(-1\) is \(-x\). So, the complete integral is \( \frac{x^2}{2} - x + C \).

Step 4: Add the constant of integration

Since it is an indefinite integral, remember to add the constant of integration \( C \) to the result. Therefore, the final answer is \( \frac{x^2}{2} - x + C \).

Key Concepts

Polynomial Long Division

Polynomial long division is a technique similar to numerical long division, but it's used to divide polynomials. It's particularly helpful when simplifying an integrand, allowing for more straightforward integration.
To start, look at each part of the problem. In this case, you're dividing the polynomial \(x^2 - 1\) by \(x + 1\).

• Divide the leading terms: Take the highest degree term from your dividend \(x^2\) and divide it by the highest degree term of the divisor \(x\). This gives \(x\).
• Multiply and subtract: Multiply \(x\) by \(x + 1\), resulting in \(x^2 + x\). Subtract this from \(x^2 - 1\), leading to a new expression \(-x - 1\).
• Repeat the process: Take \(-x\), divide by \(x\), which gives \(-1\). Multiply \(-1\) by \(x + 1\) resulting in \(-x - 1\). Subtract to get \(0\), concluding the division.

Therefore, \(\frac{x^2 - 1}{x + 1}\) simplifies to \(x - 1\), which makes the integration process smoother. This step ensures the integrand is in its simplest form for the subsequent steps of the integration.

Power Rule of Integration

The power rule of integration is a foundational tool in calculus that provides an easy way to find antiderivatives. It states that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.
This rule is applied when dealing with terms in the form of powers of \(x\). In our exercise, the term \(x\) fits this form easily.

• Integrating \(x\): The power of \(x\) is 1, so adding 1 to the power gives \(2\). The antiderivative becomes \(\frac{x^2}{2}\).
• Integrating constants: Remember, when integrating a constant like \(-1\), think of it as \(-1 \cdot x^0\). Its antiderivative is simply \(-x\), since the integral of a constant \(a\) is \(ax + C\).

These applications of the power rule enable you to write your answer as \(\frac{x^2}{2} - x + C\). The simplicity of this rule helps turn complex-looking integrals into manageable expressions.

Constant of Integration

The constant of integration, denoted by \(C\), is crucial for finding the general solution of an indefinite integral. When integrating, you solve for a family of functions that differ by a constant.
This constant arises because differentiation loses information—a derivative of a constant is zero—so when finding an antiderivative, that lost information is compensated by adding \(C\).

• Why add \(C\)?: Without \(C\), your solution might not encompass all possible functions that the original function could have represented.
• Represents an infinite family: Every possible antiderivative includes some constant, signifying the many functions that share the same derivative.

In our integral, \(\frac{x^2}{2} - x + C\), the \(C\) completes the integration. It ensures the solution is as general as possible for \(\int (x - 1) \, dx\). Ignoring \(C\) could lead to incorrect or incomplete solutions when further applying this integral solution.