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Chapter 7: Problem 25

Evaluate the integral. $$\int \frac{2 x^{2}-10 x+4}{(x+1)(x-3)^{2}} d x$$

Short Answer

Expert verified
The integral evaluates to \(2 \ln|x+1| - \frac{4}{x-3} + C\).

Step by step solution

01

Decompose into Partial Fractions

First, decompose the integrand \( \frac{2x^2 - 10x + 4}{(x+1)(x-3)^2} \) into partial fractions. We assume a form:\[\frac{2x^2 - 10x + 4}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}\]Multiply through by the denominator \((x+1)(x-3)^2\) to eliminate the denominators.
02

Set Up the Equation

Expand the equation:\[2x^2 - 10x + 4 = A(x-3)^2 + B(x+1)(x-3) + C(x+1)\]Now, simplify and equate coefficients for powers of \(x\) to solve for \(A\), \(B\), and \(C\).
03

Solve for Coefficients

1. Equate coefficients for \(x^2\), \(x\), and the constant term.2. Solve the system of equations to find \(A = 2\), \(B = 0\), \(C = -4\).The partial fraction decomposition is:\[\frac{2}{x+1} + \frac{0}{x-3} + \frac{-4}{(x-3)^2} = \frac{2}{x+1} - \frac{4}{(x-3)^2}\]
04

Integrate the Individual Terms

Now integrate each term separately:1. \( \int \frac{2}{x+1} \, dx = 2 \ln|x+1| + C_1 \)2. \( \int \frac{-4}{(x-3)^2} \, dx = 4 \cdot \frac{1}{x-3} + C_2 \)Combine these results.
05

Write the Final Answer

The integral evaluates to:\[2 \ln|x+1| - \frac{4}{x-3} + C\]where \(C\) is a constant of integration.

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Key Concepts

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

Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate or differentiate. Consider the expression in our original exercise:
  • We have \(\frac{2x^2 - 10x + 4}{(x+1)(x-3)^2}\).
  • The first step in partial fraction decomposition is to express it as a sum of simpler fractions.
The form chosen for our decomposition was: \(\frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}\).This setup allows us to solve for the constants \(A\), \(B\), and \(C\) by eliminating denominators and matching coefficients.
  • Eliminate the denominators by multiplying through by the common denominator \( (x+1)(x-3)^2 \).
  • This results in an equation we can expand and simplify.
  • By matching coefficients of corresponding \(x\) powers, we establish equations to solve for \(A\), \(B\), and \(C\).
The ultimate goal of this process is to transform a challenging integral into the sum of simpler integrals.
Definite Integrals
Definite integrals calculate the exact area under a curve between two specific values along the x-axis, known as bounds. In the context of definite integrals:
  • We assess an integral from a lower limit \(a\) to an upper limit \(b\).
  • The notation for definite integrals is \( \int_a^b f(x) \, dx \).
  • This integral yields a number representing the net area under the curve \(f(x)\) from \(x = a\) to \(x = b\).
Definite integrals differ from indefinite integrals, which include a constant of integration and do not have bounds. Thus, they provide an exact numeric result, helping us solve problems related to physics, engineering, and many other scientific fields.
Indefinite Integrals
Indefinite integrals represent a family of functions and incorporate an arbitrary constant since they do not have sold bounds like definite integrals. In solving our exercise, we evaluated indefinite integrals:
  • The indefinite integral of a function \(f(x)\) is written as \( \int f(x) \, dx \).
  • This integral results in a function \(F(x) + C\), where \(F(x)\) is the antiderivative of \(f(x)\) and \(C\) is the constant of integration.
In our example:
  • We found that \( \int \frac{2}{x+1} \, dx = 2 \ln|x+1| + C_1 \).
  • Similarly, \( \int \frac{-4}{(x-3)^2} \, dx = 4 \cdot \frac{1}{x-3} + C_2 \).
The constant \(C\) is essential as it represents an entire set of functions. It reminds us that there are infinitely many antiderivatives for a given function. This makes indefinite integrals vital for understanding the general behavior and properties of functions.

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