Problem 43 - Calculus Single Variable

Question

Simplify the integrand before integrating by parts. $$ \int x^{3} \ln (1 / x) d x $$

Step-by-Step Solution

Answer

The simplified integral is \(\frac{x^4}{4} \ln(1/x) + \frac{x^4}{16} + C\).

Step 1: Recognize the Form for Integration by Parts

For \int u \, dv, the integration by parts formula is \int u \, dv = uv - \int v \, du. Identify which part of the integrand should be \(u\) and \(dv\). In this integrand, set \(u = \ln(1/x)\) and \(dv = x^3 \, dx\).

Step 2: Differentiate and Integrate the Parts

Differentiate \(u\): \ du = -\frac{1}{x} \, dx \Integrate \(dv\): \ v = \frac{x^4}{4} \These steps provide the necessary parts for the formula.

Step 3: Apply the Integration by Parts Formula

Substitute these into the formula: \[ \int x^3 \ln(1/x) \, dx = \frac{x^4}{4} \cdot \ln(1/x) - \int \frac{x^4}{4} \cdot \left(-\frac{1}{x}\right) \, dx\] This simplifies to: \[ \frac{x^4}{4} \ln(1/x) + \frac{1}{4} \int x^3 \, dx\]

Step 4: Simplify the Integral

The remaining integral is basic: \ \int x^3 \, dx = \frac{x^4}{4} \Substitute back: \[ \frac{x^4}{4} \ln(1/x) + \frac{1}{4} \cdot \frac{x^4}{4}\] Which equals: \[ \frac{x^4}{4} \ln(1/x) + \frac{x^4}{16}\]

Step 5: Combine and Conclude

Combine the terms to get the final result: \[ \frac{x^4}{4} \ln(1/x) + \frac{x^4}{16} + C\] Where \(C\) is the constant of integration.

Key Concepts

Integral Simplification

When solving an integral, it can often be beneficial to simplify the integrand before proceeding with more complex integration techniques like integration by parts. This step can potentially save a lot of time and effort by reducing the integral to a simpler form.
For example, in the integral \( \int x^{3} \ln (1 / x) \, dx \), directly attacking this with integration by parts may seem daunting at first. However, recognizing simplifications such as \( \ln(1/x) = -\ln(x) \) can streamline the process.
Each term in the equation should be observed closely to see if there are any mathematical properties, such as logarithmic identities or algebraic simplifications, that can be applied. In this way, complexity is reduced, making the integral easier to manage.

Differentiation

Differentiation plays a crucial role when applying the integration by parts technique. The formula \( \int u \, dv = uv - \int v \, du \) requires you to differentiate one part of the integrand.
In the given exercise, we selected \( u = \ln(1/x) \). The differentiation process here is made easier by using the chain rule, leading to \( du = -\frac{1}{x} \, dx \).
This step is vital as it underpins the substitution into the integration by parts formula. Accurate differentiation ensures the rest of the integral is solved correctly, so it's important to double-check this step for accuracy.

Basic Integrals

Understanding basic integrals is essential for solving more complex problems using techniques like integration by parts. Basic integrals are simpler forms that you should be familiar with, such as \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).
In the exercise, we needed to integrate \( dv = x^3 \, dx \). This is a straightforward application of the power rule for integration, resulting in \( v = \frac{x^4}{4} \).
Remembering these basic integral forms makes evaluating the integral of \( dv \) and calculating the final solution more straightforward and efficient. Basic integrals serve as stepping stones throughout the solution process.

Constant of Integration

Whenever you integrate, remember to include the constant of integration, denoted as \( C \). This constant accounts for any constant terms that could have been present before the indefinite integration.
In the final solution, after applying integration by parts and simplifying, we ended with \[\frac{x^4}{4} \ln(1/x) + \frac{x^4}{16} + C\].
Without the constant \( C \), the solution may miss other possible solutions, since indefinite integrals have infinitely many antiderivatives differing only by a constant. Even in seemingly straightforward exercises, never neglect to add this constant at the end of your calculation; it represents the general solution for the integral.