Problem 6 - Thomas Calculus

Question

The integrals in Exercises \(1-44\) are in no particular order. Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate. When necessary, use a substitution to reduce it to a standard form. $$ \int \frac{d x}{x-\sqrt{x}} $$

Step-by-Step Solution

Answer

\(2 \ln \left| 1 - \frac{1}{\sqrt{x}} \right| + C.\)

Step 1: Simplify the Integrand

Note that the integrand is \( \frac{1}{x - \sqrt{x}} \). To simplify this expression, consider using the substitution \( u = \sqrt{x} \). Then, \( x = u^2 \), and \( dx = 2u \, du \). This leads to the transformation of the integrand and the differential: \( \frac{1}{x - \sqrt{x}} = \frac{1}{u^2 - u} \) and \( dx = 2u \, du \).

Step 2: Substitute and Adjust the Integral

Substituting \( u = \sqrt{x} \) into the integral, we have:\[\int \frac{1}{x - \sqrt{x}} \, dx = \int \frac{1}{u(u-1)} \, 2u \, du = 2 \int \frac{1}{u-1} \, du.\]

Step 3: Simplify Using Partial Fraction Decomposition

The expression \( \frac{1}{u^2 - u} \) can be decomposed using partial fractions. Since the denominator factors as \( u(u-1) \), the partial fraction decomposition is:\[\frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1}.\]Solving for \( A \) and \( B \) gives:\[A(u-1) + Bu = 1.\]By setting \( u = 0 \), \( A = -1 \). By setting \( u = 1 \), \( B = 1 \). Thus, \( \frac{1}{u(u-1)} = \frac{1}{u} - \frac{1}{u-1}. \)

Step 4: Integrate Using Decomposition

Substitute back into the integral:\[2 \int \left( \frac{1}{u-1} - \frac{1}{u} \right) du = 2 \left( \ln |u-1| - \ln |u| \right).\]This simplifies to:\[2 \ln \left| \frac{u-1}{u} \right| = 2 \ln \left| \frac{\sqrt{x} - 1}{\sqrt{x}} \right|.\]

Step 5: Back-substitute the Original Variable

Now, substitute \( u = \sqrt{x} \) back into the expression:\[2 \ln \left| \frac{\sqrt{x} - 1}{\sqrt{x}} \right| = 2 \ln \left| 1 - \frac{1}{\sqrt{x}} \right|.\] Thus, the final result is \[2 \ln \left| 1 - \frac{1}{\sqrt{x}} \right| + C,\]where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a fantastic tool in integration that can simplify complex integrals. The idea here is to transform the integral into a form that is easier to solve. For example, consider the function with a square root term like in our original problem.

• Substitution can change variables, making the integral easier to handle.
• In the problem, we used the substitution \( u = \sqrt{x} \), transforming the integrand into a simpler algebraic form.
• As a result, differential \( dx \) needs to be transformed into terms of \( du \), hence \( dx = 2u \, du \).

This converts the difficult original integral into a simpler one, which is then easier to evaluate using standard techniques. The key to this method is picking the right substitution that simplifies the integrand.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique that helps break down complex rational expressions into simpler fractions. Thus, it becomes easier to integrate.

• An expression like \( \frac{1}{u(u-1)} \) is decomposed into simpler fractions: \( \frac{A}{u} + \frac{B}{u-1} \).
• Finding constants \( A \) and \( B \) involves solving linear equations by strategically choosing values of \( u \).
• For instance, by letting \( u = 0 \), we find \( A = -1 \), and by setting \( u = 1 \), determined \( B = 1 \).

Once the original complex fraction is expressed in parts, integrating becomes much more straightforward, using basic integration rules on simpler fractions.

Trigonometric Identities

While this specific integral did not require trigonometric identities, they're indeed valuable in integration. They are essential in transforming integrals that involve trigonometric functions into more manageable forms.

• Identities such as \( \sin^2\theta + \cos^2\theta = 1 \) are often used to simplify expressions.
• They can help convert back and forth between products and sums using formulas like the angle sum identities.
• These identities become crucial especially in integrals involving trigonometric products or squared functions.

This concept, though not directly applied in the provided exercise, is part of a broader toolkit for simplifying integrals, crucial in other problems involving trigonometric elements.

Simplifying Integrals

Simplifying integrals is a fundamental step in solving complex problems. It involves a combination of algebraic manipulation and selection of suitable methods.

• Algebraic simplification may involve combining like terms or splitting fractions using algebraic identities.
• The goal is to convert the integral into a form that matches known formulas or is simple enough for basic integration techniques like substitution or partial fractions to be applied.
• In the example given, by handling the integrand \( \frac{1}{x - \sqrt{x}} \), the integral was restructured into a simpler integral \( \frac{1}{u(u-1)} \), which could then be easily decomposed.

By simplifying the integral first, solving it becomes less taxing and less error-prone. This step is crucial for efficiency and accuracy in integration.