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Chapter 7: Problem 18

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

Short Answer

Expert verified
The evaluated integral is \(\ln|x-1| + C\).

Step by step solution

01

Factor the Denominator

First, observe the denominator of the integrand, \(x^2 - 3x + 2\), and factor it into simpler terms. This quadratic expression can be factored as \((x-1)(x-2)\).
02

Use Partial Fraction Decomposition

Express the integrand as a sum of partial fractions. We can write:\[\frac{x^2}{x^2 - 3x + 2} = \frac{x^2}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\]Multiply through by the denominator \((x-1)(x-2)\) to solve for \(A\) and \(B\):\[x^2 = A(x-2) + B(x-1)\]
03

Solve for A and B

Expand and rearrange the equation:\[x^2 = Ax - 2A + Bx - B = (A+B)x - (2A+B)\]Equate the coefficients to find:1. \(A + B = 1\) for the coefficients of \(x\).2. \(-2A - B = 0\) for the constant term.Solve these simultaneous equations to find \(A=1\) and \(B=0\).
04

Change the Integrand Using Partial Fractions

Substitute the values of \(A\) and \(B\) back into the partial fraction decomposition:\[\frac{x^2}{(x-1)(x-2)} = \frac{1}{x-1} + 0\times\frac{1}{x-2} = \frac{1}{x-1}\]Thus, the integral simplifies to:\[\int \frac{1}{x-1} \, dx\]
05

Integrate the Simplified Expression

Now, integrate the simpler expression:\[\int \frac{1}{x-1} \, dx = \ln|x-1| + C\]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.

Partial Fraction Decomposition
Partial fraction decomposition is a powerful technique used in integral calculus to split a complex rational expression into simpler fractions. This method is essential for integrating functions that would otherwise be difficult to tackle. In this exercise, we decomposed \[ \frac{x^2}{x^2 - 3x + 2} \] into simpler partial fractions.
  • The aim is to express a given fraction as a sum of simpler fractions, where each has a simpler denominator.
  • Partial fraction decomposition relies on the factorization of the denominator into simpler polynomial components.
  • For instance, once the expression is separated into partial fractions, the integration process becomes straightforward.
This technique streamlines complex integrations, illustrating how algebraic manipulation can simplify calculus problems.
Factoring Polynomials
Factoring polynomials is a fundamental algebraic operation needed before employing partial fraction decomposition. This involves breaking down a polynomial into the product of simpler polynomials. In our case, the quadratic polynomial in the denominator, \( x^2 - 3x + 2 \), was factored into its components:
  • Observe the polynomial and look for factors that satisfy the equation. These factors will be in the form of smaller degree polynomials.
  • For a quadratic polynomial like \( x^2 - 3x + 2 \), we found that it factors into \((x-1)(x-2)\).
Factoring makes it possible for us to apply the partial fraction decomposition, turning a challenging integration into a more manageable one.
Integration Techniques
Once we simplify the integrand using partial fraction decomposition, the next step involves integrating the simpler expression. Integration techniques are crucial in solving definite and indefinite integrals, allowing us to find the area under a curve among other applications.
  • After decomposing the function, we were left to integrate \(\frac{1}{x-1}\), which is a much simpler task.
  • The integral of \(\frac{1}{x-1}\) is \(\ln|x-1| + C\), where \(C\) represents the constant of integration.
This task demonstrates how breaking down complex expressions makes it easier to apply integration rules, leading to solutions that are more straightforward and approachable for students.

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Most popular questions from this chapter

Use Simpson's rule approximation \(S_{10}\) to approximate the length of the curve over the stated interval. Express your answers to at least four decimal places. y=x^{-2} \text { from } x=1 \text { to } x=2

Approximate the integral using Simpson's rule \(S_{10}\) and compare your answer to that produced by a calculating utility with a numerical integration capability. Express your answers to at least four decimal places. $$ \int_{-1}^{2} x \sqrt{2+x^{3}} d x $$

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{x}{(x+3)^{1 / 5}} d x $$

Determine whether the statement is true or false. Explain your answer. If \(f(x)\) is concave down on the interval \((a, b)\), then the trapezoidal approximation \(T_{n}\) underestimates \(\int_{a}^{b} f(x) d x\)

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 \(v_{\mathrm{rms}}=\sqrt{3 R T / M}\)
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