Problem 66 - Applied Calculus

Question

Find the indefinite integrals. $$\int\left(x^{2}+\frac{1}{x}\right) d x$$

Step-by-Step Solution

Answer

\( \frac{x^3}{3} + \ln |x| + C \)

Step 1: Split the Integral

Start by breaking the integral into separate parts:\[ \int \left( x^2 + \frac{1}{x} \right) dx = \int x^2 \, dx + \int \frac{1}{x} \, dx \]

Step 2: Integrate Each Part

Integrate each term separately. For the first term, \( \int x^2 \, dx \), the integral is calculated using the power rule for integration. For the second term, \( \int \frac{1}{x} \, dx \), use the basic integral rule.For \( x^2 \):\[ \int x^2 \, dx = \frac{x^{3}}{3} + C_1 \]For \( \frac{1}{x} \):\[ \int \frac{1}{x} \, dx = \ln |x| + C_2 \]

Step 3: Combine the Results

Once both integrals are computed, combine them into a single expression. Introduce a constant \( C \) to represent the sum of the constants \( C_1 \) and \( C_2 \):\[ \int \left( x^2 + \frac{1}{x} \right) dx = \frac{x^{3}}{3} + \ln |x| + C \]

Step 4: Final Answer Verification

Make sure to verify each integrated part matches expected calculus rules and ensure constants are correctly generalized. The integral of power and logarithmic functions should align with standard operations.

Key Concepts

Indefinite Integrals

Indefinite integrals are a cornerstone of integral calculus. They are used to find functions whose derivatives give us the integrand. In simpler terms, indefinite integrals help us "reverse" the process of differentiation. An indefinite integral is expressed with the integral symbol \( \int \), a function \( f(x) \), and the differential \( dx \). Importantly, an indefinite integral is not limited by bounds, hence the name. The result includes a "constant of integration" \( C \). This constant accounts for any constant that might disappear during differentiation. The general form of an indefinite integral is:

• \( \int f(x) \, dx = F(x) + C \)

Where \( F(x) \) is the antiderivative of \( f(x) \). In our exercise, after integrating each term separately, we found the solution \( \frac{x^{3}}{3} + \ln |x| + C \). In this expression, \( C \) is critical as it signifies that indefinitely many antiderivatives exist for the given function.

Integration Techniques

Various integration techniques help solve different kinds of integrals. Similar to tools in a toolbox, each technique has its optimal application. In our exercise, we used the method of splitting the integral, which simplifies the process by addressing each part separately.One simple but valuable technique is breaking down a complex integral into smaller, manageable parts. This technique was evident when we applied the rule:

• \( \int (x^2 + \frac{1}{x}) \, dx = \int x^2 \, dx + \int \frac{1}{x} \, dx \)

Once broken down, we applied other techniques like substitution (where possible), or standard rules directly applicable to each simpler integral.Choosing the right integration technique can greatly simplify the process, leading to correct and efficient solutions. In practical terms, recognizing the nature of each term in an integral is crucial to applying the most appropriate technique for integration.

Power Rule for Integration

One of the most basic and powerful techniques for integrating polynomials is the power rule for integration. This rule simplifies integrals of the form \( x^n \), where \( n \) is any real number except \(-1\). The power rule is elegantly simple:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) where \( n eq -1 \)

In the exercise, the power rule was used to integrate \( x^2 \) easy-peasy. We simply add 1 to the exponent and divide by the new exponent:

• \( \int x^2 \, dx = \frac{x^{3}}{3} + C_1 \)

This demonstrates how straightforward the power rule is for polynomial terms. However, be cautious when \( n = -1 \), as it requires a different approach (typically leading to a logarithmic function). In summary, the power rule for integration is a fundamental tool, essential for efficiently solving a wide array of integral calculus problems.