Problem 8 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{d x}{x-\sqrt{x}} $$

Step-by-Step Solution

Answer

The integral is \( 2 \ln{|\sqrt{x} - 1|} + C \).

Step 1: Identify the Substitution

Identify a suitable substitution to simplify the integral. Here, let \( u = \sqrt{x} \). Then \( x = u^2 \) and \( dx = 2u \, du \).

Step 2: Apply the Substitution

Substitute \( x = u^2 \) and \( dx = 2u \, du \) into the integral. The integral becomes \( \int \frac{2u \, du}{u^2 - u} \).

Step 3: Simplify the Integrand

Simplify the integrand: \( \frac{2u}{u^2 - u} = \frac{2u}{u(u - 1)} = \frac{2}{u - 1} \).

Step 4: Integrate the Simplified Form

Evaluate the integral \( \int \frac{2}{u - 1} \, du \), which is \( 2 \ln{|u - 1|} + C \).

Step 5: Back-Substitute

Replace \( u \) with \( \sqrt{x} \) to get the integral in terms of \( x \). This yields \( 2 \ln{|\sqrt{x} - 1|} + C \).

Step 6: Final Expression

Present the final expression: The evaluated integral is \( 2 \ln{|\sqrt{x} - 1|} + C \).

Key Concepts

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in calculus. An indefinite integral, such as \[ \int f(x) \, dx \] represents a family of antiderivatives. This means it accounts for all possible functions that, when differentiated, give you the integrand. Typically, indefinite integrals include a constant of integration, represented as \[ + C \] because the differentiation of a constant is zero, leaving the possibility that some constant could have been part of the original function.

On the other hand, a definite integral calculates the area under a curve between two specified points of the independent variable. When you see something like \[ \int_{a}^{b} f(x) \, dx \] you are calculating the net area between the curve \[ f(x) \] and the x-axis from \[ x = a \] to \[ x = b \].
Despite these differences, both types of integrals rely heavily on similar techniques for simplification and problem-solving, such as substitution. This helps in finding simpler expressions to work with.

Simplification of Integrands

Integrands often need to be simplified to make integration manageable. This simplification can come in various forms, especially when dealing with more complex functions.

Substitution is a common method for simplifying an integrand. It involves finding a substitution for a variable that simplifies the integral into a more recognizable form. For example, in our exercise, substituting \( u = \sqrt{x} \) transformed \( x \) into \( u^2 \) and the differential \( dx \) into \( 2u \, du \). This made the integral easier to handle.

After substitution, the integrand \( \frac{d x}{x-\sqrt{x}} \) was simplified to \( \frac{2}{u-1} \) .
This new form is much simpler and more straightforward to integrate. Simplification can also involve taking common factors out, breaking complex fractions, or using algebraic identities.

Successful simplification can make a challenging integration task a breeze! It's a crucial skill for any calculus student.

Calculus Problem Solving Steps

Solving calculus problems systematically can make complex tasks manageable. Here's a general approach using substitution in integrals:

• **Identify the Problem**: Start by recognizing the form of the integrand and consider potential substitution candidates. Look for patterns like nested functions or expressions where a derivative is present.


• **Decide on a Substitution**: Choose a substitution that reduces complexity. For instance, if you notice an expression like \( \sqrt{x} \) in the integrand, try setting it as \( u \) , turning the problem into one involving simpler differentials.


• **Convert and Simplify**: Use the substitution to rewrite the integral in terms of the new variable. Simplify the integrand to a known or standard form that is easier to integrate.


• **Evaluate the Integral**: Perform the integration on the simplified form using known integral rules. Remember, this will usually result in an antiderivative expressed with your substitution variable.


• **Back-Substitute**: Revert your substitution to express the antiderivative in terms of the original variable. This is crucial for the final solution to relate back to the original problem.


• **Include the Constant (Indefinite Integrals)**: Don't forget the constant of integration \( C \) for indefinite integrals. It's a frequent detail that completes the solution.

These steps serve as a roadmap to follow for integral problems. The process encourages a structured approach, breaking down complex problems into manageable tasks, which is key to mastering calculus.