Problem 50 - 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

Rewrite the integral by splitting it into two separate integral expressions. This can be expressed as: \[\int \left(x^2 + \frac{1}{x}\right) dx = \int x^2 \, dx + \int \frac{1}{x} \, dx\]This allows us to focus on solving each integral separately.

Step 2: Solve the first integral \(\int x^2 \, dx\)

To integrate \(x^2\), use the power rule for integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Here, \(n=2\).- Apply the rule: \[\int x^2 \, dx = \frac{x^{3}}{3} + C_1\]- This represents the antiderivative of \(x^2\).

Step 3: Solve the second integral \(\int \frac{1}{x} \, dx\)

To integrate \(\frac{1}{x}\), recall that the antiderivative is the natural logarithm of the absolute value of \(x\). The formula is:\[\int \frac{1}{x} \, dx = \ln|x| + C_2\]This represents the antiderivative of \(\frac{1}{x}\).

Step 4: Combine results

Combine the results from Step 2 and Step 3 to find the indefinite integral of the original expression. Remember to combine the constant terms as they represent any constant of integration:\[\int \left(x^2 + \frac{1}{x}\right) dx = \frac{x^3}{3} + \ln|x| + C\]where \(C = C_1 + C_2\) is the overall constant of integration.

Key Concepts

Power Rule of Integration

The power rule of integration is a fundamental concept when dealing with polynomial functions. It allows us to find the antiderivative of any polynomial expression, which is a crucial step in solving indefinite integrals. If you have a term like \(x^n\), the power rule states:

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

Here, \(n\) can be any real number except -1, since dividing by zero would be undefined. The term \(\frac{x^{n+1}}{n+1}\) is what you get after increasing the exponent by one and then dividing by the new exponent. Find the constant of integration \(C\), which we'll explore more later, after integrating all terms.
For example, in our exercise, when integrating \(x^2\), we apply this rule with \(n=2\), resulting in \(\frac{x^{3}}{3} + C_1\). This outcome represents the antiderivative for \(x^2\). Always remember to apply this rule term by term when integrating polynomials.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), plays a significant role when dealing with the integral of the reciprocal function \(\frac{1}{x}\). It is derived from the fact that the derivative of \(\ln|x|\) is \(\frac{1}{x}\), making it the perfect antiderivative.

• \( \int \frac{1}{x} \, dx = \ln|x| + C \)

This form accounts for all values of \(x\) except zero, which captures the behavior of \(\frac{1}{x}\) accurately over its domain excluding zero. It is crucial to include the absolute value notation \(|x|\) to handle both positive and negative values of \(x\), unless the problem specifies \(x\) as positive only.
In the exercise, finding the antiderivative for \(\frac{1}{x}\) involves recognizing that it simplifies to \(\ln|x| + C_2\), which emphasizes its reliance on the natural logarithm to undo the derivative function.

Constant of Integration

When we integrate a function to find its antiderivative, the result is not a single function but a family of functions. This is because integration is the inverse operation of differentiation, and differentiation of any constant is zero. Therefore, we add a constant of integration, denoted \(C\), to account for all possible constant shifts in the antiderivative.

• \( \int f(x) \, dx = F(x) + C \)

This constant is essential because different functions with the same derivative could differ by a constant. In indefinite integrals, each processed term gains its constant of integration, yet when the terms are combined, these constants merge into a single \(C\).
In the context of the step-by-step solution, finalizing the integration of \(x^2\) and \(\frac{1}{x}\) separately, we combine the constants \(C_1\) and \(C_2\) into \(C = C_1 + C_2\) to represent any uniform adjustment to the integrated function.