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Chapter 6: Problem 30

Use the table of integrals in this section to find the indefinite integral. $$ \int \frac{2 x}{(1-3 x)^{2}} d x $$

Short Answer

Expert verified
The indefinite integral of \( \int \frac{2 x}{(1-3 x)^{2}} dx \) is \( - \frac{2}{3} (\ln |1-3x| + C) \).

Step by step solution

01

Identify the substitution

Looking at the integral \[ \int \frac{2 x}{(1-3 x)^{2}} dx \] , it appears that a good choice for substitution would be \( u = 1-3x \), as it simplifies the denominator and the derivative of this function appears in the numerator.
02

Substitute and integrate

First, find the derivative of \( u = 1-3x \) which is \( du = -3 dx \) . To make the substitution, solve this equation for \( dx \) to get \( dx = - \frac{1}{3} du \). Now replace \( 1-3x \) with \( u \) and \( dx \) with \( - \frac{1}{3} du \) in the original integral. The integral becomes \( - \frac{2}{3} \int \frac{u^{-2}}{ u } du \). Now the integral can be simplified to \( - \frac{2}{3} \int u^{-1} du \). At this step, the standard integral \(\int u^{-1} du = \ln |u|\) can be applied.
03

Calculate the integral and substitute back

Using the standard integral, calculate the integral to get \( - \frac{2}{3} (\ln |u| + C) \), where \( C \) is the constant of integration. Lastly, substitute back \( u = 1-3x \) to get \( - \frac{2}{3} (\ln |1-3x| + C) \) as the solution to the integral.

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Key Concepts

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

Integration Techniques
Integration stands as a fundamental component of calculus, closely tied to the concept of finding areas or the antiderivatives of functions.
One of the most essential integration techniques is the method of substitution, often referred to as u-substitution. This method is particularly useful for integrating composite functions.
Another technique involves integrating by parts which is the integral counterpart of the product rule for differentiation. Partial fraction decomposition is another valuable technique for dealing with rational functions.
The choice of technique largely depends on the form of the function being integrated.
U-Substitution
U-substitution is a technique used in integration that involves substituting part of the integrand with a new variable, typically denoted as 'u'.
This method simplifies the integral by identifying a segment within the integrand, the derivative of which is also present. When an appropriate substitution is made, the integral often becomes more straightforward to solve. It's akin to translating a difficult expression into a simpler 'language' with the goal of making the process of integration more manageable.
Integral of Rational Functions
Rational functions are ratios of polynomials. The integration of rational functions can be tricky, but there are strategies to handle them, such as long division when the degree of the numerator is higher than the degree of the denominator, or partial fraction decomposition when the denominator can be factored.
Recognizing and applying these strategies allow for breaking down complex-looking integrals into simpler parts that can be more readily integrated.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is an important function in calculus, particularly in integration. Its properties are often used when integrating functions that resemble the derivative of the logarithm function.
For instance, the integral of \( 1/x \) is \( \ln |x| \), plus a constant of integration. Having a firm grasp of the natural logarithm and its relationship to e, the base of natural logarithms, is crucial for solving many integral problems involving exponential and logarithmic functions.

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Most popular questions from this chapter

Probability Use a program similar to the Simpson's Rule program on page 454 with \(n=6\) to approximate the indicated normal probability. The standard normal probability density function is \(f(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}\). If \(x\) is chosen at random from a population with this density, then the probability that \(x\) lies in the interval \([a, b]\) is \(P(a \leq x \leq b)=\int_{a}^{b} f(x) d x\). $$ P(0 \leq x \leq 1.5) $$

Present Value Use a program similar to the Simpson's Rule program on page 454 with \(n=8\) to approximate the present value 454 the income \(c(t)\) over \(t_{1}\) years at the given annual interest rate \(r .\) Then use the integration capabilities of a graphing utility to approximate the present value. Compare the results. (Present value is defined in Section \(6.1 .)\) $$ c(t)=6000+200 \sqrt{t}, r=7 \%, t_{1}=4 $$

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n .\) (Round your answers to three significant digits.) $$ \int_{0}^{1} \sqrt{1-x^{2}} d x, n=4 $$

Probability Use a program similar to the Simpson's Rule program on page 454 with \(n=6\) to approximate the indicated normal probability. The standard normal probability density function is \(f(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}\). If \(x\) is chosen at random from a population with this density, then the probability that \(x\) lies in the interval \([a, b]\) is \(P(a \leq x \leq b)=\int_{a}^{b} f(x) d x\). $$ P(0 \leq x \leq 1) $$

Capitalized cost Find the capitalized cost \(C\) of an asset (a) for \(n=5\) years, (b) for \(n=10\) years, and (c) forever. The capitalized cost is given by $$ C=C_{0}+\int_{0}^{n} c(t) e^{-r t} d t $$ where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance (measured in dollars). [Hint: For part (c), see Exercises \(35-38 .]\) $$ C_{0}=\$ 650,000, c(t)=25,000(1+0.08 t), r=12 \% $$
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