Problem 34 - Applied Calculus

Question

Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. $$ \int \frac{x^{2}-x}{2 x^{3}-3 x^{2}} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{6} \ln |2x^3 - 3x^2| + C \).

Step 1: Identify a Suitable Substitution

We aim to simplify the integral \( \int \frac{x^{2}-x}{2 x^{3}-3 x^{2}} dx \) by using substitution. Look for parts of the integral where differentiation would match other parts. Here, recognize that the derivative of \(2x^3 - 3x^2\) could involve \(x^2 - x\). Calculating it gives: \( \frac{d}{dx}(2x^3 - 3x^2) = 6x^2 - 6x\), which simplifies to \(6(x^2-x)\). Thus, use the substitution: \(u = 2x^3 - 3x^2\).

Step 2: Differentiate and Simplify the Substitution

Differentiate \(u = 2x^3 - 3x^2\). Calculate \( \frac{du}{dx} = 6x^2 - 6x = 6(x^2-x) \). This implies \( du = 6(x^2-x) dx \). We have \( dx = \frac{du}{6(x^2-x)} \). Substitute these into the integral.

Step 3: Substitute and Simplify the Integral

Replace \(dx\) in the original integral with \(\frac{du}{6(x^2-x)}\) and adjust the rest of the terms based on substitution: \( \int \frac{x^{2}-x}{u} \cdot \frac{1}{6(x^2-x)} du = \int \frac{1}{6u} du\). Now, integrate \( \frac{1}{6u} \).

Step 4: Integrate the Simplified Integral

The integral \( \int \frac{1}{6u} du \) is straightforward: \( \frac{1}{6} \int \frac{1}{u} du = \frac{1}{6} \ln |u| + C \). Substitute back \(u = 2x^3 - 3x^2\) to obtain the result in terms of \(x\).

Step 5: Final Answer in Terms of Original Variables

Substitute back \(u\) in terms of \(x\): \( \frac{1}{6} \ln |2x^3 - 3x^2| + C \). Hence, the indefinite integral is \( \frac{1}{6} \ln |2x^3 - 3x^2| + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus for simplifying complex integrals. It involves transforming a difficult integral into an easier one by changing the variable of integration. To use this method, we need to identify a part of the integral that can be replaced with a simpler expression. This often involves spotting a function within the integral that resembles a derivative of another function. In our exercise, we used the substitution \( u = 2x^3 - 3x^2 \). The reason for choosing this substitution is that the derivative of the expression \( 2x^3 - 3x^2 \) closely matches \( x^2 - x \), making it a suitable candidate for simplification. After substitution, the integral becomes much easier to evaluate, leading us directly to the answer without complicated algebraic manipulation.

Integration Techniques

Integration techniques are methods used to find the integral of a function, especially when the function appears complex or does not have a straightforward antiderivative. There are several techniques, among which substitution is commonly used. For integrals that involve a composite of functions, substitution helps to reduce the integral into a more manageable form:

• Identify potential substitutions: Look for expressions whose derivative appears somewhere in the integral.
• Differentiate and substitute: Once a suitable expression is chosen, replace the variables to transform the integral.
• Proceed with simplified integration: Often, after substitution, the integral becomes a standard form which can be readily solved.
• Back-substitute: After integration, replace the substitution variable back in terms of the original variable.

These steps streamline the integration process, enabling you to solve integrals that might initially seem challenging.

Calculus Problem Solving

Calculus problem solving requires a blend of analytical skills, creativity, and an understanding of mathematical techniques. When tackling an integration problem, it's essential to break it into smaller, more manageable steps. Begin by identifying the type of functions you are dealing with, which allows you to choose the appropriate solving method, such as substitution. For example, recognizing that an integral is a composite function hints that substitution might simplify it. Additionally, it's crucial to maintain a logical progression by checking each step as you go along. This might involve verifying calculations, ensuring substitutions are correct, and confirming that the transformed integral is solvable. Post-solving, review the solution in terms of the original problem. This final check ensures that your work is correct and that the results make sense in the context of the problem. Effective problem-solving in calculus is about methodically applying the right techniques and verifying the solutions at each stage.