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Chapter 8: Problem 34

$$\text {Evaluate the following integrals.}$$ $$\int \frac{3 x^{2}+4 x-6}{x^{2}-3 x+2} d x$$

Short Answer

Expert verified
Answer: The result of the integral is \(3x - 5\ln|x-1| + 2\ln|x-2| + C\), where \(C\) is the constant of integration.

Step by step solution

01

Perform long division

Divide the numerator \((3x^2 + 4x - 6)\) by the denominator \((x^2 - 3x + 2)\) to obtain a polynomial and a remainder that can be written as a fraction of the original denominator: $$\frac{3x^2 + 4x - 6}{x^2 - 3x + 2} = 3 - \frac{7x - 12}{(x^2 - 3x + 2)}$$
02

Factor the denominator

Factor the denominator \((x^2 - 3x + 2)\) to get: $$(x^2 - 3x + 2) = (x-1)(x-2)$$
03

Perform partial fraction decomposition

Express the remaining fraction as the sum of simpler fractions with the original factors of the denominator as their denominators: $$\frac{7x - 12}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$$ To find the constants \(A\) and \(B\), set the right-hand side sum equal to the original fraction and clear the denominators: $$7x - 12 = A(x-2) + B(x-1)$$
04

Solve for constants A and B

To find \(A\) and \(B\), substitute \(x=1\) and \(x=2\) into the equation: For \(x=1\): $$7(1) - 12 = A(1-2) \Rightarrow A = 5$$ For \(x=2\): $$7(2) - 12 = B(2-1) \Rightarrow B = 2$$ Now we have: $$\frac{7x - 12}{(x-1)(x-2)} = \frac{5}{x-1} + \frac{2}{x-2}$$
05

Integrate the decomposed fraction

Now integrate the separated fractions: $$\int (3 - \frac{5}{x-1} + \frac{2}{x-2}) \, dx = 3\int dx - 5\int \frac{1}{x-1} dx + 2\int \frac{1}{x-2} dx$$
06

Complete the integration

Integrate each term: $$3x - 5\ln|x-1| + 2\ln|x-2| + C$$ So the final result is: $$\int \frac{3x^2 + 4x - 6}{x^2 - 3x + 2} dx = 3x - 5\ln|x-1| + 2\ln|x-2| + C$$

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Key Concepts

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

Definite Integrals in Calculus
Definite integrals in calculus are a fundamental concept, often representing the area under a curve or a physical quantity like work or probability. When evaluating definite integrals, the key is to first find the indefinite integral or the antiderivative of the function, and then apply the limits of integration.

For example, in the exercise we're looking at, the indefinite integral of \( \frac{3x^2 + 4x - 6}{x^2 - 3x + 2} \) was found using partial fraction decomposition and polynomial long division. If the question had been posed as a definite integral with limits, say from \(a\) to \(b\), we would then plug these limits into the result to obtain the final answer. It's crucial that when we evaluate the antiderivative at the upper and lower limits, we subtract the latter from the former, symbolically: \( F(b) - F(a) \), where \( F(x) \) is the antiderivative found.
Polynomial Long Division
Polynomial long division is a technique similar to long division with numbers, and it's used when the degree of the numerator (the top part of a fraction) is greater than or equal to the degree of the denominator (the bottom part). The goal is to rewrite the fraction in a simpler form.

In our exercise, the numerator \(3x^2 + 4x - 6\) has the same degree as the denominator \(x^2 - 3x + 2\), so we performed polynomial long division to obtain a polynomial plus a remainder fraction. This step simplifies the original integrand, making it easier to handle, especially when we move on to partial fraction decomposition. It is a crucial step as it converts complex rational expressions into simpler ones that can be integrated more straightforwardly.
Integration by Substitution
Integration by substitution, also known as u-substitution, is a method for finding integrals that resemble the reverse chain rule. This technique involves substituting part of the integrand with a new variable \(u\) to simplify the integral. It's very useful when dealing with composite functions or when the integrand's derivative is present elsewhere in the expression.

Although integration by substitution wasn't directly used in the exercise provided, it's important to recognize when this method can be applied. For instance, if after partial fraction decomposition we obtained a term like \( \frac{1}{(2x-4)^2} \) in the integrand, we could let \( u = 2x-4 \) to simplify the integration process. Remembering to substitute back for \(x\) after integrating with respect to \(u\) is equally critical to arriving at the correct solution.

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Most popular questions from this chapter

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=t \rightarrow F(s)=\frac{1}{s^{2}}$$

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}(\) coulombs ) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\).

A remarkable integral \(1 \mathrm{t}\) is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4},\) for all real numbers \(m\). a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2,1\) 3/2, and 2, and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m\).

Use numerical methods or a calculator to approximate the following integrals as closely as possible. The exact value of each integral is given. $$\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x=-\frac{\pi \ln 2}{2}$$

The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system. $$\int x \sin ^{-1} 2 x d x$$
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