Problem 31 - Applied Calculus

Question

Find each indefinite integral. \(\int \frac{z^{3}+z}{z^{2}} d z\)

Step-by-Step Solution

Answer

\( \frac{z^2}{2} + \ln |z| + C \)

Step 1: Simplify the Integrand

Start by simplifying the expression inside the integral. Divide each term in the numerator by the denominator: \[ \frac{z^3}{z^2} + \frac{z}{z^2} \Rightarrow z + z^{-1} \]Thus, the integral changes to: \[ \int (z + z^{-1}) \, dz \]

Step 2: Integrate Each Term Separately

Integrate each term of the expression separately:1. The integral of \( z \) with respect to \( z \) is \( \frac{z^2}{2} \).2. The integral of \( z^{-1} \) with respect to \( z \) is \( \ln |z| \). Hence, \[ \int (z + z^{-1}) \, dz = \frac{z^2}{2} + \ln |z| \]

Step 3: Add the Constant of Integration

Since the integral is indefinite, add a constant of integration \( C \): \[ \frac{z^2}{2} + \ln |z| + C \]

Step 4: Conclusion

The indefinite integral of the given expression is \[ \frac{z^2}{2} + \ln |z| + C \]

Key Concepts

Integration Techniques

When it comes to solving indefinite integrals, choosing the right integration technique is crucial. In this exercise, we began by simplifying the integrand. Simplifying often makes the rest of the problem much easier.
For the problem \( \int \frac{z^3 + z}{z^2} \, dz \), we simplified the expression by dividing each term in the numerator by the denominator. This step is not always necessary, but it can save a lot of time and confusion by converting the integrand into a form that is easier to integrate.
Once simplified to \( z + z^{-1} \), the integral involves basic rules:

• The power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) when \( n eq -1 \)
• The logarithmic rule: \( \int \frac{1}{x} \, dx = \ln|x| + C \)

Mastering these foundational rules and identifying opportunities to simplify can significantly enhance your problem-solving skills in calculus problems.

Calculus Problems

Calculus can often feel like a daunting subject, with its numerous types of problems and solutions. One key aspect is identifying and solving indefinite integrals, which represent a family of functions, plus a constant of integration \( C \).
In the example \( \int (z + z^{-1}) \, dz \), we encounter both polynomial and reciprocal terms. This combination showcases how various types of terms can be tackled using basic integrative rules.

• The polynomial term \( z \) becomes \( \frac{z^2}{2} \)
• The reciprocal term \( z^{-1} \) translates to \( \ln |z| \)

Adding a constant \( C \) is important in indefinite integrals, as it accounts for all possible vertical shifts of the antiderivative on a graph. Remember, the goal is to derive a function whose derivative matches the original function provided.

Simplifying Expressions

Simplifying expressions is a fundamental step in solving many mathematical problems, including calculus integrals. By reducing a complex fraction, \( \frac{z^3+z}{z^2} \), simplifying transforms it to \( z + z^{-1} \) which is easier to integrate term-by-term.
This technique reduces potential errors and makes the computation straightforward.

• Simplification involves dividing each term separately: \( \frac{z^3}{z^2} = z \) and \( \frac{z}{z^2} = z^{-1} \).
• The result is a simpler sum of terms that can be approached individually in the integration process.

Practical simplification not only aids in solving the problem more efficiently but also hones one's ability to manipulate algebraic expressions effectively. This is vital across many branches of mathematics and related fields.