Problem 38 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

The integral is \( \frac{x^2}{2} + x + C \).

Step 1: Simplify the Integrand

First, simplify the integrand \( \frac{x^2 - 1}{x - 1} \). Notice that the numerator \( x^2 - 1 \) is a difference of squares, which can be factored as \( (x + 1)(x - 1) \). Dividing \( (x + 1)(x - 1) \) by \( x - 1 \), we find that the expression simplifies to \( x + 1 \) since the term \( x - 1 \) cancels out.

Step 2: Write the Simplified Integral

After simplification, the integral becomes: \[ \int (x + 1) \, dx \]

Step 3: Integrate Each Term

Integrate \( x \) and \( 1 \) separately.- The integral of \( x \) is \( \frac{x^2}{2} \).- The integral of \( 1 \) is \( x \).Thus, the integral is: \[ \frac{x^2}{2} + x + C \] where \( C \) is the constant of integration.

Key Concepts

Factorization Demystified

Factorization is a crucial technique in algebra used for simplifying expressions. It involves breaking down complex terms into simpler ones that can easily be managed or further manipulated.
In our given exercise, the expression \( x^2 - 1 \) is known as a difference of squares because it can be expressed as a subtraction between two squared terms: \( x^2 \) and \( 1^2 \). This expression can be factored into the product \( (x + 1)(x - 1) \).
By identifying patterns such as these, you can transform expressions into a more workable form. For instance, knowing how to factor a difference of squares allows for quick simplification, essential for moving on to integration. Recognizing immediately that \( x^2 - 1 \) can be factored saves time and effort while simplifying the integrand.

Simplification of Expressions

Simplification is the process of reducing expressions to their simplest form without changing their value. It's like cleaning up a messy room where everything becomes easy to find and recognize.
After factorization in our exercise, we have the expression \( \frac{(x+1)(x-1)}{x-1} \). The term \( x-1 \) in the numerator and denominator cancel each other out, leaving us with just \( x + 1 \).

Through simplification, the integral becomes less daunting. It's shifted from a complex fraction to a basic polynomial, paving the way for straightforward integration. Effective simplification isn't just about algebraic cleaning; it sets the stage for applying other powerful mathematical techniques efficiently.

Mastering Integration Techniques

Integration is one of the core concepts in calculus, akin to finding the sum of infinitely small quantities. When dealing with indefinite integrals, such as \( \int (x + 1) \, dx \), the process requires a clear understanding of how to integrate different terms.

For the term \( x \), we apply the power rule of integration: increase the power by one and divide by the new exponent. So, \( \int x \, dx = \frac{x^2}{2} \). For the constant \( 1 \), its integral is simply \( x \) because the derivative of \( x \) gives us \( 1 \).

Finally, remember to include the constant of integration, \( C \), because indefinite integrals represent a family of functions. These integration techniques transform expressions to provide a clear mathematical expression of accumulated continuous quantities.