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Chapter 7: Problem 21

Evaluate the integral. \(\int \frac{x^{5}+x^{2}+2}{x^{3}-x} d x\)

Short Answer

Expert verified
\(\frac{x^3}{3} + \frac{x^2}{2} + x - 2 \ln|x| + 2 \ln|x-1| + C\)

Step by step solution

01

Simplify the Integrand

First, notice that the denominator can be factored as follows:\[ x^3 - x = x(x^2 - 1) = x(x-1)(x+1) \]This factoring helps us later when we decompose the fraction.
02

Perform Polynomial Long Division

Since the degree of the numerator (5) is greater than the degree of the denominator (3), perform polynomial long division. Divide \(x^5 + x^2 + 2\) by \(x^{3} - x\). This yields:\[x^2 + x + 1\]with a remainder of \(2x + 2\). Thus, the integral becomes:\[ \int (x^2 + x + 1) \, dx + \int \frac{2x + 2}{x^3 - x} \, dx\]
03

Integrate the Polynomial Part

Integrate the polynomial part \(x^2 + x + 1\):\[ \int(x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C_1\]
04

Decompose the Fraction

Now, perform partial fraction decomposition on the remaining integral:\[\frac{2x + 2}{x(x-1)(x+1)}\]Assume:\[\frac{2x + 2}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\]Solve for \(A\), \(B\), and \(C\) by equating coefficients.
05

Solve for Coefficients

Set up the equation:\[ 2x + 2 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)\]Substituting suitable values for \(x\), we find:- Let \(x=0:\) \[ 2 = -A \Rightarrow A = -2 \]- Let \(x=1:\) \[ 4 = 2B \Rightarrow B = 2 \]- Let \(x=-1:\) \[ 0 = 2C \Rightarrow C = 0 \]
06

Integrate the Decomposed Fraction

Plug the obtained values \(A = -2\), \(B = 2\), and \(C = 0\) back into the partial fraction form:\[\int \left(\frac{-2}{x} + \frac{2}{x-1}\right) \, dx\]Integrate term by term:\[-2 \int \frac{1}{x} \, dx + 2 \int \frac{1}{x-1} \, dx = -2 \ln|x| + 2 \ln|x-1| + C_2\]
07

Combine Results

Put everything together:\[ \int \frac{x^5 + x^2 + 2}{x^3 - x} \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x - 2 \ln|x| + 2 \ln|x-1| + C\]where \(C\) is the constant of integration.

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Key Concepts

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

Polynomial Long Division
Polynomial long division is a technique used to divide a polynomial by another polynomial of lower degree. In the context of integration, it helps us simplify an integrand when the degree of the numerator is higher than the degree of the denominator. For example, given
  • numerator: \(x^5 + x^2 + 2\)
  • denominator: \(x^3 - x\)
We perform the division like we do with numbers, dividing the highest degree term in the numerator by the highest degree term in the denominator. This produces a new polynomial, simplifying the problem into:
  • \(x^2 + x + 1\)
with a remainder, which helps break down complex rational expressions into manageable parts for integration.
Partial Fraction Decomposition
Partial fraction decomposition is a method to express a rational function as a sum of simpler fractions. This is especially useful in integration when faced with complex denominators, like in our problem where remainder is
  • \(\frac{2x + 2}{x(x-1)(x+1)}\)
We write it as a sum of fractions with unknown coefficients:
  • \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)
We then solve for \(A\), \(B\), and \(C\) by plugging in suitable values for \(x\) to simplify the equations. This makes the integration process easier by transforming the original function into a sum of fractions we can easily integrate one by one.
Logarithmic Integration
Logarithmic integration is a technique used when integrating functions of the form \(\int \frac{1}{x} \, dx\), which results in a logarithmic expression. In our exercise, once partial fraction decomposition is applied, the integration includes terms like
  • \(\int \frac{-2}{x} \, dx\)
  • \(\int \frac{2}{x-1} \, dx\)
These are straightforward to integrate, leading to logarithmic expressions:
  • \(-2 \ln|x|\)
  • \(2 \ln|x-1|\)
It鈥檚 important to remember to include the absolute value in logarithmic functions to handle any potential negative values of \(x\), ensuring the function remains defined for all values where \(x eq 0\) and \(x eq 1\).
Indefinite Integrals
Indefinite integrals represent a family of functions and include an arbitrary constant, often denoted as + C. They provide a way to find the antiderivative of a function, representing all possible antiderivatives. After you've performed operations like polynomial long division and partial fraction decomposition, you solve for the antiderivative of each component.In our example, once the integral is broken down, each part is integrated separately:
  • \(\int(x^2 + x + 1) \, dx\)
  • \(\int \left(\frac{-2}{x} + \frac{2}{x-1}\right) \, dx\)
These are then combined into a single expression with a general constant + C. This constant reflects the integration's indefinite nature, reminding us that there could be multiple solutions differing by a constant.

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Most popular questions from this chapter

The average speed, \(\bar{v}\), of the molecules of an ideal gas is given by $$ \bar{v}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{+\infty} v^{3} e^{-M v^{2} /(2 R T)} d v $$ and the root-mean-square speed, \(v_{\mathrm{rms}}\), by $$ v_{\mathrm{rms}}^{2}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{+\infty} v^{4} e^{-M v^{2} /(2 R T)} d v $$ where \(v\) is the molecular speed, \(T\) is the gas temperature, \(M\) is the molecular weight of the gas, and \(R\) is the gas constant. (a) Use a CAS to show that $$ \int_{0}^{+\infty} x^{3} e^{-a^{2} x^{2}} d x=\frac{1}{2 a^{4}}, \quad a>0 $$ and use this result to show that \(\bar{v}=\sqrt{8 R T /(\pi M)}\). (b) Use a CAS to show that $$ \int_{0}^{+\infty} x^{4} e^{-a^{2} x^{2}} d x=\frac{3 \sqrt{\pi}}{8 a^{5}}, \quad a>0 $$ and use this result to show that \(v_{\mathrm{rms}}=\sqrt{3 R T / M}\)

Some integrals that can be evaluated by hand cannot be evaluated by all computer algebra systems. Evaluate the integral by hand, and determine if it can be evaluated on your CAS. $$ \int \frac{x^{3}}{\sqrt{1-x^{8}}} d x $$

Engineers want to construct a straight and level road \(600 \mathrm{ft}\) long and \(75 \mathrm{ft}\) wide by making a vertical cut through an intervening hill (see the accompanying figure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying figure. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road.] $$ \begin{array}{cc} \hline \begin{array}{l} \text { HORIZONTAL } \\ \text { DISTANCE } x \text { (ft) } \end{array} & \begin{array}{c} \text { HEIGHT } \\ h(\mathrm{ft}) \end{array} \\ \hline 0 & 0 \\ 100 & 7 \\ 200 & 16 \\ 300 & 24 \\ 400 & 25 \\ 500 & 16 \\ 600 & 0 \end{array} $$

Medication can be administered to a patient using a variety of methods. For a given method, let \(c(t)\) denote the concentration of medication in the patient's bloodstream (measured in \(\mathrm{mg} / \mathrm{L}) t\) hours after the dose is given. The area under the curve \(c=c(t)\) over the time interval \([0,+\infty)\) indicates the "availability" of the medication for the patient's body. Determine which method provides the greater availability. Method \(1: c_{1}(t)=5\left(e^{-0.2 t}-e^{-t}\right)\) Method 2: \(c_{2}(t)=4\left(e^{-0.2 t}-e^{-3 t}\right)\)

The integral \(\int \tan ^{4} x \sec ^{5} x d x\) is equivalent to one whose integrand is a polynomial in \(\sec x\).
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