Problem 32 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int x^{-3}(x+1) d x $$

Step-by-Step Solution

Answer

The most general antiderivative is \(-\frac{1}{x} - \frac{1}{2x^2} + C\).

Step 1: Express the Integral

The exercise requires finding the antiderivative of the function \( x^{-3}(x+1) \). First, expand the expression inside the integral: \[ \int x^{-3}(x+1) \, dx = \int (x^{-3} \cdot x + x^{-3} \cdot 1) \, dx = \int (x^{-2} + x^{-3}) \, dx \] Now, we have a simpler form to integrate.

Step 2: Integrate Term by Term

We integrate each term separately. The integration formula for \( x^n \) is \( \frac{x^{n+1}}{n+1} \) for \( n eq -1 \).1. For \( \int x^{-2} \, dx \), integrate to get \( -x^{-1} = -\frac{1}{x} \).2. For \( \int x^{-3} \, dx \), integrate to get \( -\frac{1}{2}x^{-2} = -\frac{1}{2x^2} \). Combine these results to get the general antiderivative:\[ -\frac{1}{x} - \frac{1}{2x^2} + C \] where \( C \) is the constant of integration.

Step 3: Differentiate to Check

Differentiate the antiderivative found in Step 2 to verify the result. Differentiate \( -\frac{1}{x} - \frac{1}{2x^2} + C \):1. The derivative of \( -\frac{1}{x} \) is \( x^{-2} \).2. The derivative of \( -\frac{1}{2x^2} \) is \( x^{-3} \).Combine the derivatives: \[ x^{-2} + x^{-3} \] This matches the original integrand \( x^{-2} + x^{-3} \), confirming our solution is correct.

Key Concepts

Antiderivatives

When we talk about antiderivatives, we are essentially looking for a function whose derivative is the given function. An antiderivative is a way to reverse the process of differentiation. This process is also referred to as finding the indefinite integral of a function.

In our exercise, we needed to find the most general antiderivative for the function \( x^{-3}(x+1) \). By expanding the expression, it turns into the simpler form \( x^{-2} + x^{-3} \). This makes it easier to manage, as we can now focus on integrating each term separately.

Once we find the antiderivatives of each term, we combine them to form the general solution. It's essential to include a constant of integration, \( C \), in our final answer because indefinite integrals can differ by a constant, due to the constant disappearing during differentiation.

Integration Techniques

Integration is a fundamental concept in calculus, and there are various techniques to tackle different types of integrals. In this exercise, we used a straightforward technique: integrating term by term.

A useful formula when dealing with power functions is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \), applicable when \( n eq -1 \). This formula becomes particularly handy when the integrand is a polynomial or can be expressed as a sum of power functions. For our function, after simplifying \( x^{-3}(x+1) \), the main task was to apply this formula to each resulting term separately:

• Turn \( x^{-2} \) into its antiderivative, which yields \( -\frac{1}{x} \).
• Similarly, \( x^{-3} \) becomes \( -\frac{1}{2x^2} \) as its antiderivative.

By systematically applying integration techniques, we obtained the complete antiderivative: \( -\frac{1}{x} - \frac{1}{2x^2} + C \). This approach is crucial for managing more extensive and nuanced integrals efficiently.

Differentiation Check

After finding an antiderivative, it is always a good practice to check your work by differentiating the result. Differentiation will help verify the correctness of your antiderivative. If the derivative of your solution matches the original integrand, your solution is correct.

In our exercise, we differentiated the solution \(-\frac{1}{x} - \frac{1}{2x^2} + C\). Here's the differentiation step-by-step:

• The derivative of \(-\frac{1}{x}\) is \(x^{-2}\).
• The derivative of \(-\frac{1}{2x^2}\) is \(x^{-3}\).

Adding these derivatives together gives us \(x^{-2} + x^{-3}\), which matches the original expression under the integral sign \(x^{-2} + x^{-3}\).

This confirmation step is beneficial, as it reinforces your understanding of both integration and differentiation and ensures the accuracy of your mathematical operations. It's a reliable method to catch any errors that might have occurred during integration.