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Chapter 8: Problem 14

Evaluate the following integrals. $$\int \frac{x^{-2}+x^{-3}}{x^{-1}+16 x^{-3}} d x$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{x^{-2}+x^{-3}}{x^{-1}+16 x^{-3}} dx$$. Solution: The integral can be evaluated as $$\int \frac{x^{-2}+x^{-3}}{x^{-1}+16 x^{-3}} dx = x^4+C_1x^3 - \frac{15}{2} \ln |x^2+16| + C_2$$, where \(C_1\) and \(C_2\) are constants of integration.

Step by step solution

01

Rewrite the integral using positive exponents

Let's rewrite the given integral using positive exponents: $$\int \frac{x^{-2}+x^{-3}}{x^{-1}+16 x^{-3}} dx = \int \frac{\frac{1}{x^2}+\frac{1}{x^3}}{\frac{1}{x}+16\frac{1}{x^3}} dx$$ Apply reverse power rule for integration to the given integral: $$= x^3 \int \frac{\frac{1}{x^2}+\frac{1}{x^3}}{\frac{1}{x}+16\frac{1}{x^3}} dx$$
02

Simplify the integral once again

Now, multiply the numerator and the denominator by \(x^3\) to simplify: $$= x^3 \int \frac{x+x^2}{x^2+16} dx$$
03

Perform partial fraction decomposition

To perform partial fraction decomposition, we use $$\frac{x+x^2}{x^2+16} = A + \frac{Bx}{x^2+16}$$ where A and B are constants. To find the constants, we write: $$x+x^2 = A(x^2+16) + Bx$$ Comparing coefficients, we have: $$x^2: 1 = A \Rightarrow A = 1$$ $$x: 1 = B + 16A \Rightarrow B = 1 - 16 = -15$$ So we rewrite the integral as: $$x^3 \int \left(1 - \frac{15x}{x^2+16} \right) dx$$
04

Integrate individual terms

Now, we can integrate each term separately: $$x^3 \int (1 - \frac{15x}{x^2+16}) dx = x^3 ( \int 1 dx - 15 \int \frac{x}{x^2+16} dx)$$ Now, integrate the first term using the reverse power rule: $$x^3 (\int 1 dx) = x^3(x+C_1) = x^4+C_1x^3$$ For the second term, we perform a u-substitution: $$u = x^2+16 \Rightarrow du = 2x dx$$ $$15 \int \frac{x}{x^2+16} dx = \frac{15}{2} \int \frac{du}{u}$$ Integrating the second term yields: $$\frac{15}{2} \int \frac{du}{u} = \frac{15}{2} \ln |u| + C_2 = \frac{15}{2} \ln |x^2+16| + C_2$$ Now combine both terms: $$x^3 ( \int 1 dx - 15 \int \frac{x}{x^2+16} dx) = x^4+C_1x^3 - \frac{15}{2} \ln |x^2+16| + C_2$$
05

Final answer

Combining all the steps, the definite integral is evaluated as: $$\int \frac{x^{-2}+x^{-3}}{x^{-1}+16 x^{-3}} dx = x^4+C_1x^3 - \frac{15}{2} \ln |x^2+16| + C_2$$ where \(C_1\) and \(C_2\) are constants 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 in calculus to break down rational expressions into simpler fractions that are easier to integrate. When dealing with a fraction where the degrees of the numerator and denominator are close, partial decomposition can simplify integration by allowing you to integrate separate parts.

For example, in the given integral, we have:
  • Numerator: \(x + x^2\)
  • Denominator: \(x^2 + 16\)
By expressing this as \(A + \frac{Bx}{x^2 + 16}\), we transform the problem into one of solving for constants \(A\) and \(B\). This is done by equating the expressions:
  • \(x + x^2 = A(x^2 + 16) + Bx\)
From here, comparing coefficients helps us find \(A = 1\) and \(B = -15\). Now the integral can be separated into simpler parts:
  • \(x^3 \int \left(1 - \frac{15x}{x^2 + 16}\right) dx\)
This restructuring forms the basis for performing integration in separate, simpler steps.
U-Substitution
U-substitution is a helpful method to simplify integration, particularly when dealing with composite functions. It involves replacing a part of the integral with a single variable \(u\), making the integral easier.In the original problem, the second term \(\int \frac{15x}{x^2+16} dx\) requires u-substitution.
  • Let \(u = x^2 + 16\)
  • Then, \(du = 2x \, dx\)
  • Rewriting gives \(dx = \frac{du}{2x}\)
By substituting, the integral becomes:
  • \(\frac{15}{2} \int \frac{du}{u}\)
This now simplifies to \(\frac{15}{2} \ln |u| + C_2\), which can be back-substituted for \(u\) as \(\frac{15}{2} \ln |x^2+16| + C_2\).
U-substitution transforms the original complex functions into basic integrals of \(\ln\), making computations manageable.
Definite Integral
Definite integrals are used to calculate the net area under a curve within a specific interval. Although this exercise ends with an indefinite integral, the techniques like those used here convert indefinite integrals into definite ones by evaluating limits when provided.

The given exercise illustrates performing indefinite integration through several steps and highlights the use of constants \(C_1\) and \(C_2\). These constants signify the presence of an indefinite integral:
  • \(x^4 + C_1x^3 - \frac{15}{2} \ln |x^2+16| + C_2\)
When dealing with definite integrals, these constants vanish due to applying limits. Nevertheless, understanding these steps, including partial fraction decomposition and u-substitution, lays the groundwork for tackling definite integrals systematically.
Though indefinite integrals provide a general solution, converting them into definite integrals can help solve problems concerning area or accumulated change over an interval.

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Most popular questions from this chapter

Determine whether the following integrals converge or diverge. $$\int_{3}^{\infty} \frac{d x}{\ln x}(\text { Hint: } \ln x \leq x .)$$

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower. ) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R .\) Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\)

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. More than one integration method can be used to evaluate \(\int \frac{d x}{1-x^{2}}\) b. Using the substitution \(u=\sqrt[3]{x}\) in \(\int \sin \sqrt[3]{x} d x\) leads to \(\int 3 u^{2} \sin u d u\) c. The most efficient way to evaluate \(\int \tan 3 x \sec ^{2} 3 x d x\) is to first rewrite the integrand in terms of \(\sin 3 x\) and \(\cos 3 x\) d. Using the substitution \(u=\tan x\) in \(\int \frac{\tan ^{2} x}{\tan x-1} d x\) leads to \(\int \frac{u^{2}}{u-1} d u\)

Another Simpson's Rule formula Another Simpson's Rule formula is \(S(2 n)=\frac{2 M(n)+T(n)}{3},\) for \(n \geq 1 .\) Use this rule to estimate \(\int_{1}^{e} \frac{1}{x} d x\) using \(n=10\) subintervals.

Let \(L(c)\) be the length of the parabola \(f(x)=x^{2}\) from \(x=0\) to \(x=c,\) where \(c \geq 0\) is a constant. a. Find an expression for \(L\) b. Is \(L\) concave up or concave down on \([0, \infty) ?\) c. Show that as \(c\) becomes large and positive, the are length function increases as \(c^{2}\); that is, \(L(c) \approx k c^{2},\) where \(k\) is a constant.
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