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Chapter 8: Problem 64

Evaluate the following integrals. $$\int \frac{\ln a x}{x} d x, a \neq 0$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{\ln(ax)}{x} dx\). Answer: \(\int \frac{\ln(ax)}{x} dx = (\ln(ax) - C_1) \cdot \ln|x| - \int \frac{\ln |x|}{x} dx + C\), where \(C_1\) and \(C\) are constants.

Step by step solution

01

Identify u and dv

We are given the integral \(\int \frac{\ln(ax)}{x} dx\). To use integration by parts, we let \(u = \ln(ax)\) and \(dv = \frac{1}{x} dx\).
02

Differentiate u and integrate dv

Differentiate \(u\) with respect to \(x\) to find \(du\): $$ du = \frac{d(\ln(ax))}{dx} = \frac{a}{ax} dx = \frac{1}{x} dx $$ Integrate \(dv\) with respect to \(x\) to find \(v\): $$ v = \int \frac{1}{x} dx = \ln|x| + C_1 $$ where \(C_1\) is an arbitrary constant.
03

Use integration by parts formula

Now, we can use the integration by parts formula: \(\int u dv = uv - \int v du\). Plugging in the values for \(u\), \(v\), \(du\), and \(dv\), we get: $$ \int \frac{\ln(ax)}{x} dx = \left(\ln(ax)\right)(\ln|x| + C_1) - \int(\ln|x| + C_1) \frac{1}{x} dx $$
04

Simplify and integrate the remaining integral

Expand the integral: $$ \int \frac{\ln(ax)}{x} dx = \ln(ax) \cdot \ln|x| - C_1\ln(ax) - \int \left(\frac{\ln|x|}{x} + \frac{C_1}{x}\right) dx $$ Now, we need to integrate the remaining terms: $$ \int \left(\frac{\ln a x}{x} + \frac{C_1}{x}\right) dx = \int \frac{\ln |x|}{x} dx + C_1 \int \frac{1}{x}dx $$ Note that \(\int \frac{1}{x} dx = \ln|x| + C_2\), where \(C_2\) is another arbitrary constant. Therefore, $$ \int \left(\frac{\ln a x}{x} + \frac{C_1}{x}\right) dx = \int \frac{\ln |x|}{x} dx + C_1 \ln|x|+ C_2 $$ To find the integral \(\int \frac{\ln |x|}{x} dx\), we can use integration by parts once more, but this time the integral becomes the same as the original one. Therefore, we can't find an elementary function for this integral. So, our answer will be in terms of this integral.
05

Write the final result

Substitute the result from Step 4 back into the expression from Step 3: $$ \int \frac{\ln(ax)}{x} dx = \ln(ax) \cdot \ln|x| - C_1\ln(ax) - \left(\int \frac{\ln |x|}{x} dx + C_1 \ln|x| + C_2\right) $$ Combine the constants \(C_1\) and \(C_2\) into a single constant \(C\) and simplify: $$ \int \frac{\ln(ax)}{x} dx = (\ln(ax) - C_1) \cdot \ln|x| - \int \frac{\ln |x|}{x} dx + C $$ This is the final result for the given integral.

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Key Concepts

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

Logarithmic Integration
Logarithmic integration is a technique that often involves integrals where a natural logarithm term, such as \( \ln(x) \), is present. It becomes particularly useful when using integration by parts, as it allows us to handle different functions that multiply with such logarithmic terms. In the exercise at hand, we encounter the natural logarithm, \( \ln(ax) \), divided by \( x \). By appropriately choosing which part of the expression to differentiate and integrate, logarithmic functions can simplify integrals that otherwise seem challenging. The key is to use integration by parts to break down the function into more manageable parts.Integration by parts relies on the formula:\[ \int u \, dv = uv - \int v \, du \], where you choose \( u \) and \( dv \) strategically. Here, selecting \( u = \ln(ax) \) allows us to easily differentiate it and make the calculation more tractable.
Definite and Indefinite Integrals
Integrals are essential components of calculus, mainly classified as definite and indefinite. An indefinite integral, often denoted by \( \int f(x) \, dx \), represents the general form of the antiderivative function without specific bounds. The result includes a constant, typically written as \( C \), to account for all potential antiderivatives.
  • Indefinite integrals represent families of functions, each differing by a constant.
  • Definite integrals have set limits of integration, giving a precise area under the curve.
In our exercise, the focus is on indefinite integrals, where solutions generally present one or more constants (e.g., \( C_1, C_2 \)). These constants are valuable because they represent a family of solutions rather than a single curve.When dealing with complex logarithmic expressions, indefinite integrals can appear indefinite without a simple elementary result, highlighting the beauty and complexity of calculus.
Calculus Techniques
Calculus offers various methods to tackle problems, and among them, Integration by Parts is one of the most powerful techniques. It's specifically handy when you deal with complicated product integrals, such as \( \int \frac{\ln(ax)}{x} \, dx \), as seen in the problem.For effective application, always:
  • Select \( u \) and \( dv \) so that \( du \) and \( v \) simplify the problem.
  • Ensure the choice reduces complexity after applying \( \int u \, dv = uv - \int v \, du \).
Simplifying integrals that are initially complex through parts requires thoughtful selection and sometimes revisiting integration by parts to achieve the solution.Furthermore, combining integration by parts with other calculus techniques, such as logarithmic properties and simplifications, paves the way to solving intricate integrals. This combination forms the core of many calculus problems, ensuring students gain both a fundamental and applied understanding of mathematical principles.

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Most popular questions from this chapter

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-a x^{2}}\) a. Graph the Gaussian for \(a=0.5,1,\) and 2. b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.

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, \(F(x)=\frac{G M m}{x^{2}}\) 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. $$\int \frac{d x}{1-\tan ^{2} x}$$

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 \(\int \frac{u^{2}}{u-1} d u\)

Estimating error Refer to Theorem 8.1 in the following exercises. Let \(f(x)=\sqrt{\sin x}\) a. Find a Simpson's Rule approximation to \(\int_{1}^{2} \sqrt{\sin x} d x\) using \(n=20\) subintervals. b. Find an upper bound on the absolute error in the estimate found in part (a) using Theorem 8.1. (Hint: Use the fact that \(\left.\left|f^{(4)}(x)\right| \leq 1 \text { on }[1,2] .\right)\)
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