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Chapter 7: Problem 58

Evaluate the following integrals. $$\int \frac{d x}{x^{2}-6 x+34}$$

Short Answer

Expert verified
Answer: The antiderivative of the given integral is \(\arctan\left(\frac{x-3}{5}\right) + C\).

Step by step solution

01

Complete the square on the denominator

In the denominator we have \(x^2 - 6x + 34\). Completing the square on the quadratic expression, we add and subtract \((-6/2)^2 = 9\) to maintain equality: \((x^2 - 6x + 9) + 25 = (x - 3)^2 + 5^2\) Now, the integral becomes $$\int \frac{dx}{(x-3)^2 + 5^2}$$
02

Identify the standard form of the integral and use a suitable substitution

Now, we can see that this integral is in the standard form of $$\int \frac{dx}{a^2 + x^2}$$ where \(a = 5\). This form of integral has the antiderivative \(\frac{1}{a}\arctan(\frac{x}{a}) + C\). To make the substitution, let \(x - 3 = 5\tan(\theta)\), so \(dx = 5\sec^2(\theta)d\theta\) and we have $$\int \frac{5\sec^2(\theta)d\theta}{5^2+(\tan(\theta))^2\cdot5^2}$$
03

Simplify and integrate

The integral now simplifies to $$\int \frac{\sec^2(\theta)}{1+\tan^2(\theta)}d\theta$$ Using the trigonometric identity \(1+\tan^2(\theta) = \sec^2(\theta)\), the integral simplifies further: $$\int d\theta=\theta + C$$
04

Substitute back the original variables

Now, let's return back to our original substitution. Since \(x - 3 = 5\tan(\theta)\), we have \(\theta = \arctan(\frac{x-3}{5})\). Thus, the antiderivative of the given integral is $$\arctan \left(\frac{x-3}{5}\right) + C$$

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Key Concepts

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

Completing the Square
Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This simplifies the quadratic into a form that is easier to integrate. In the original exercise, we start with the quadratic polynomial in the form of:
  • \(x^2 - 6x + 34\)
To complete the square, identify the coefficient of the linear term, which is -6, divide it by 2, and then square it:
  • \((-6/2)^2 = 9\)
Add and subtract 9 inside the expression to maintain equality:
  • \((x^2 - 6x + 9) + 25\)
This results in the perfect square
  • \((x - 3)^2 + 25\)
This transformation simplifies the integration process by converting the denominator into a recognizable standard form.
Standard Integral Forms
Standard integral forms are key tools in solving integrals that conform to a certain pattern or "template". Recognizing these forms can greatly simplify the integration process. After completing the square, the integral becomes:
  • \( \int \frac{dx}{(x-3)^2 + 5^2} \)
This matches the standard form of \( \int \frac{dx}{a^2 + x^2} \), where \( a=5 \). The solution to this standard form is a widely known antiderivative:
  • \( \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C \)
Recognizing this allows us to avoid complicated integration and immediately move towards solving the integral using this well-known template.
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify the integration of functions involving square roots and certain rational expressions. In our example, following the transformation using the standard form, we make the substitution
  • \( x - 3 = 5\tan(\theta) \)
This implies:
  • \( dx = 5\sec^2(\theta)d\theta \)
Substituting these into the integral, we transform it further:
  • \( \int \frac{5\sec^2(\theta)d\theta}{25+25\tan^2(\theta)} \)
Using the trigonometric identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), the integral simplifies significantly:
  • \( \int d\theta \)
Thus, trigonometric substitution changes a complicated expression into an easily integrable form.
Antiderivative
The antiderivative, or indefinite integral, is the function that reverses the process of differentiation. It represents a family of functions whose derivatives are given by a specific function. In solving the given integral, after simplifying the expression, we arrive at:
  • \( \int d\theta = \theta + C \)
Since \( \theta \) was defined in terms of \( x \) using the substitution \( x - 3 = 5\tan(\theta) \), we find that:
  • \( \theta = \arctan \left( \frac{x - 3}{5} \right) \)
Therefore, the antiderivative of the initial integral is:
  • \( \arctan \left( \frac{x - 3}{5} \right) + C \)
This final expression represents the solution to the integral, capturing the essence of reversing differentiation.

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Most popular questions from this chapter

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}\) a. Find the work required to launch an object in terms of \(m\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18 th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2}\). For Earth to be a black hole, what would its radius need to be?

Evaluate the following integrals or state that they diverge. $$\int_{0}^{9} \frac{d x}{(x-1)^{1 / 3}}$$

Evaluating an integral without the Fundamental Theorem of Calculus Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$ \int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x $$ b. Use part (a) and the identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal 33, 4, Sep 2004).

Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). \(\int_{0}^{\pi / 2} \cos ^{9} x d x=\frac{128}{315}\)

Suppose that the rate at which a company extracts oil is given by \(r(t)=r_{0} e^{-k t},\) where \(r_{0}=10^{7}\) barrels \(/ \mathrm{yr}\) and \(k=0.005 \mathrm{yr}^{-1} .\) Suppose also the estimate of the total oil reserve is \(2 \times 10^{9}\) barrels. If the extraction continues indefinitely, will the reserve be exhausted?
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