Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{x+1}{9 x^{2}+18 x+10} d x $$

Step-by-Step Solution

Answer

\(-\frac{1}{9} \ln|3(x+1)-1| + \frac{2}{9} \ln|3(x+1)+1| + C\)

Step 1: Simplify the Fraction

First, let's simplify the given expression inside the integration. We have \( \frac{x+1}{9x^2 + 18x + 10} \).

Step 2: Complete the Square in the Denominator

The denominator is \(9x^2 + 18x + 10\). We'll complete the square to rewrite it:\[9(x^2 + 2x) + 10 = 9((x+1)^2 - 1) + 10 = 9(x+1)^2 - 9 + 10 = 9(x+1)^2 - 1.\]

Step 3: Substitute to Simplify

Let \( u = x+1 \) so that \( du = dx \). The integral becomes:\[\int \frac{u - 1}{9u^2 - 1} \, du.\]

Step 4: Partial Fraction Decomposition

Use partial fraction decomposition to express \( \frac{u-1}{9u^2 - 1} = \frac{u-1}{(3u-1)(3u+1)} \) as a sum of fractions:\[\frac{u-1}{(3u-1)(3u+1)} = \frac{A}{3u-1} + \frac{B}{3u+1}.\]Multiply through by the denominator to solve for \( A \) and \( B \):\[u - 1 = A(3u+1) + B(3u-1).\]Solve for \( A \) and \( B \).

Step 5: Solve for Constants

By comparing coefficients, we solve for \( A \) and \( B \):1. \( A(3u) + B(3u) = u \Rightarrow 3A + 3B = 1 \)2. \( A(1) - B(1) = -1 \Rightarrow A - B = -1 \).Solve these equations:- From 2: \( A = B - 1 \).- Substitute into 1: \( 3(B-1) + 3B = 1 \Rightarrow 6B - 3 = 1 \Rightarrow 6B = 4 \Rightarrow B = \frac{2}{3} \).- Substitute \( B \) into the equation for \( A \): \( A = \frac{2}{3} - 1 = -\frac{1}{3} \).

Step 6: Integrate Each Term

Substitute the values of \( A \) and \( B \) back into the partial fraction:\[\int \frac{-1/3}{3u-1} \, du + \int \frac{2/3}{3u+1} \, du.\]Integrate each:\[-\frac{1}{3} \int \frac{1}{3u-1} \, du + \frac{2}{3} \int \frac{1}{3u+1} \, du.\]The integrals are:\[-\frac{1}{3} \cdot \frac{1}{3} \ln|3u-1| + \frac{2}{3} \cdot \frac{1}{3} \ln|3u+1|.\]

Step 7: Finalize the Integration and Back-substitute

Simplify the result:\[-\frac{1}{9} \ln|3u-1| + \frac{2}{9} \ln|3u+1| + C.\]Remember that \( u = x+1 \), so our final answer is:\[-\frac{1}{9} \ln|3(x+1)-1| + \frac{2}{9} \ln|3(x+1)+1| + C.\]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to simplify the integration of rational functions. It involves breaking down a complex fraction into a sum of simpler fractions.
This makes them easier to integrate.
Given a fraction \(\frac{P(u)}{Q(u)}\) where \(Q(u)\) can be factorized into simpler linear or quadratic factors, the integral becomes manageable by expressing it as a sum:
- \(\frac{P(u)}{Q(u)} = \frac{A}{3u-1} + \frac{B}{3u+1}\)Here, \(A\) and \(B\) are constants determined by substituting back and solving equations based on equivalent expressions. For instance, solving:
\(u - 1 = A(3u+1) + B(3u-1)\)
helps find these constants.
Once the partial fractions are determined, each can be integrated separately, simplifying the overall process.

Completing the Square

Completing the square is a method used to simplify quadratic expressions, like those found in integration problems.
It rewrites these expressions in a form that is easier to work with, particularly for integration.For an expression like \(9x^2 + 18x + 10\), the completion of the square involves rearranging and factoring:
- Start with \(9(x^2 + 2x) + 10\).- Rewrite \(x^2 + 2x\) as \((x+1)^2 - 1\).
- This results in \(9(x+1)^2 - 9 + 10\), simplifying to \(9(x+1)^2 - 1\).This method changes the quadratic into \((x+1)^2\) form, making substitution and integration much easier by reducing complexity. It is a crucial step to simplify terms before further steps like substitution occur.

Substitution Method

The substitution method, also known as "u-substitution," is a common integration technique aimed at simplifying the integration process by changing variables. This is particularly useful when dealing with composite functions.
In this exercise, we let \(u = x+1\), hence \(du = dx\), modifying our integral to:
\(\int \frac{u - 1}{9u^2 - 1} \, du\).This change of variables is strategic for various reasons:
- It simplifies the algebraic expression in the integrand.
- It helps in aligning the integral with known forms or prepares it for partial fraction decomposition.
After substitution, the problem becomes more straightforward and more closely resembles a standard integral type that we already know how to solve more efficiently.

Indefinite Integrals

Indefinite integrals represent a fundamental operation in calculus where the result is a family of functions, every function differing by a constant.
This integral results in the antiderivative of a given function, expressed as \(F(x) + C\). The "+ C" represents the arbitrary constant of integration, accounting for the fact that antiderivatives are not unique.For the given problem, indefinite integral applies through sequential simplification techniques:
- After partial fractions, the integral \(\int \frac{-1/3}{3u-1} \, du\) and \(\int \frac{2/3}{3u+1} \, du\) are independently integrated.- The results from each simplified integral are combined, yielding:
\[-\frac{1}{9} \ln|3u-1| + \frac{2}{9} \ln|3u+1| + C\].
This demonstrates how indefinite integration ties various techniques together to provide a comprehensive solution.