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Chapter 5: Problem 25

First make a substitution and then use integration by parts to evaluate the integral. \(\int x \ln (1+x) d x\)

Short Answer

Expert verified
The integral evaluates to \( \frac{(1 + x)^2}{2} \ln(1 + x) - \frac{(1 + x)^2}{4} - (1 + x) \ln(1 + x) + (1 + x) + C \).

Step by step solution

01

Choose a Substitution

Let's start by making the substitution \( u = 1 + x \). As a result, the differential \( du = dx \) and \( x = u - 1 \). This will change the integral to \( \int (u - 1) \ln(u) \, du \).
02

Distribute and Re-split the Integral

Write the integral as the sum of two separate integrals: \( \int (u \ln(u) - \ln(u)) \, du \). This can be split into \( \int u \ln(u) \, du - \int \ln(u) \, du \).
03

Evaluate \( \int u \ln(u) \, du \) Using Integration by Parts

Choose \( v = \ln(u) \), which implies \( dv = \frac{1}{u} \, du \), and let \( dw = u \, du \), which means \( w = \frac{u^2}{2} \). Using integration by parts, we have \[ \int u \ln(u) \, du = \frac{u^2}{2} \ln(u) - \int \frac{u^2}{2} \frac{1}{u} \, du \]. Simplify the remaining integral to \( \frac{1}{2} \int u \, du \) and solve to get \( \frac{u^2}{2} \ln(u) - \frac{u^2}{4} + C_1 \).
04

Evaluate \( \int \ln(u) \, du \) Using Integration by Parts

Choose \( v = \ln(u) \), so \( dv = \frac{1}{u} \, du \). Let \( dw = du \), which means \( w = u \). Using integration by parts, the formula becomes \[ \int \ln(u) \, du = u \ln(u) - \int u \frac{1}{u} \, du \]. This simplifies to \( u \ln(u) - u + C_2 \).
05

Combine and Substitute Back

Combine the results from Steps 3 and 4. The integral becomes \[ \left( \frac{u^2}{2} \ln(u) - \frac{u^2}{4} \right) - \left( u \ln(u) - u \right) + C \]. Substitute back \( u = 1 + x \) to obtain \[ \frac{(1 + x)^2}{2} \ln(1 + x) - \frac{(1 + x)^2}{4} - (1 + x) \ln(1 + x) + (1 + x) + C \]. Simplify, if possible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
When faced with evaluating complex integrals, it's essential to employ various integration techniques to simplify the problem. In calculus, integration by parts and substitution are two fundamental approaches:

  • Substitution Method: This technique involves making a substitution to transform the integral into a simpler form. Often, a substitution is chosen to eliminate a complex function or simplify the expression.
  • Integration by Parts: Once the integral has been simplified, integration by parts can be employed. This method is based on the product rule for differentiation and is very useful when dealing with products of functions.
  • To perform integration by parts, we typically choose parts of the integral to differentiate and integrate, denoted as:
    • Let \(v\) be a differentiable function, such that \(dv\) is straightforward to find.
    • Let \(w\) be an integrable function from which \(dw\) can be easily derived.
This clever processing of the integral often reduces it to more manageable components, leading to a complete solution.
Substitution Method
The substitution method is a pivotal tool in integration used to simplify complex expressions. It essentially involves replacing one part of the integrand with a different variable.
  • In the given exercise, the substitution \(u = 1 + x\) was chosen. This was strategic as it simplifies the logarithm function \ln(1+x)\ into \ln(u)\.
  • We also calculate the differential \(du = dx\), which helps in rewriting the integral entirely in terms of the new variable \(u\).
  • Such substitution transforms an initially intimidating integral into a form that separates into simpler integrals or allows for direct application of other techniques, like integration by parts.

With the integral expressed in these new terms, solving it becomes more straightforward. This highlights substitution as a valuable weapon in the calculus toolkit, especially when dealing with transcendental functions like logarithms.
Definite Integrals
Definite integrals represent the accumulation of quantities and are used to find areas under curves. They have fixed upper and lower limits representing the range of integration. While this exercise discusses an indefinite integral, understanding definite integrals' concepts can still be insightful:
  • Definite integrals evaluate the net area between the curve and the x-axis, from the lower to the upper limit.
  • They yield a numerical value, offering insights into quantities like area, volume, and total accumulation within specified boundaries.
  • The computation often follows similar methods used for indefinite integrals, such as substitution and integration by parts, but requires evaluating the antiderivative at the specified limits and subtracting the results.

Definite integrals are foundational in calculus, enabling the analysis of real-world phenomena involving continuous changes across intervals.
Mathematical Problem Solving
Mathematical problem-solving in calculus often involves a blend of various techniques to arrive at a solution. This integral problem beautifully illustrates the process:
  • Break Down the Problem: Start by simplifying the integral using substitution or by recognizing patterns.
  • Choose Appropriate Methods: Apply integration techniques like integration by parts, considering the complexity and nature of functions involved.
  • Combine the Results: Once individual integrals are solved, ensure to combine them correctly, accounting for constants and any substitutions made earlier.
  • Verify and Simplify: Verify the solution by differentiating the result, ensuring it matches the original integrand, and simplifying wherever possible to obtain a clean, final expression.

This structured approach not only solves the integral but also fosters a deeper understanding, preparing you for tackling more advanced problems with confidence.

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