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Chapter 7: Problem 21

Find the integrals $$\int \frac{\ln x}{x^{2}} d x$$

Short Answer

Expert verified
\(-\frac{\ln x}{x} + \frac{1}{x} + C\)

Step by step solution

01

Recognize the Integration Technique

The integral involves a logarithmic function divided by a power of x. This suggests that integration by parts might be appropriate. Integration by parts is given by: \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) appropriately.
02

Choose Functions for Integration by Parts

Let's take \( u = \ln x \) because the derivative of \( \ln x \) simplifies the expression, and \( dv = \frac{1}{x^2} \, dx \). Then, we find \( du \) by differentiating \( u \): \( du = \frac{1}{x} \, dx \). Find \( v \) by integrating \( dv \): \( v = -\frac{1}{x} \).
03

Apply the Integration by Parts Formula

Substitute \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula: \[ \int \frac{\ln x}{x^2} \, dx = \ln x \left(-\frac{1}{x}\right) - \int -\frac{1}{x} \cdot \frac{1}{x} \, dx \]. Simplify this to: \[ -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx \].
04

Simplify and Integrate the Remaining Integral

The remaining integral is \( \int \frac{1}{x^2} \, dx \). This simplifies to \( \int x^{-2} \, dx \). Integrate using the power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), finding \( \int x^{-2} \, dx = -x^{-1} = -\frac{1}{x} \).
05

Combine Results

Combine the integrated parts: \[ -\frac{\ln x}{x} - \left(-\frac{1}{x}\right) + C = -\frac{\ln x}{x} + \frac{1}{x} + C \], 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.

integration by parts
Integration by parts is a powerful technique used to solve integrals involving products of functions. The method is based on the product rule of differentiation and is formulated as \( \int u \, dv = uv - \int v \, du \). Here, "\( u \)" and "\( dv \)" are carefully chosen parts of the original integral.
  • We decide "\( u \)" to be a part of the integral that becomes simpler when differentiated, whereas "\( dv \)" is a component that remains manageable upon integration.
  • After selecting \( u \) and \( dv \), we determine \( du \) and \( v \). \( u \) is differentiated to find \( du \), and \( dv \) is integrated to find \( v \).
Effective use of this technique requires practice in choosing the right parts for "\( u \)" and "\( dv \)". In the given exercise, selecting "\( u = \ln x \)" and "\( dv = \frac{1}{x^2} \; dx \)" simplifies the integral, allowing us to proceed and solve the problem efficiently.
logarithmic integration
Logarithmic integration often involves integrals with logarithmic functions like \( \ln x \). In cases where such functions merge with other types of functions, integration by parts can be beneficial.
  • Choosing "\( u = \ln x \)" is common because its derivative "\( du = \frac{1}{x} \; dx \)" simplifies integrals significantly.
  • This results in expressions that are more straightforward to integrate or differentiate.
Whenever you encounter an integral involving a logarithmic function, consider exploring logarithmic properties and integration by parts as techniques to simplify the problem. This approach was effectively applied in integrating \( \int \frac{\ln x}{x^2} \; dx \), simplifying the task by handling the logarithmic component separately.
power rule in integration
The power rule in integration is a straightforward and fundamental rule that applies to functions in the form of \( x^n \). It states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). This rule simplifies integrals by offering a quick step for problems with power functions.
  • In this exercise, after applying integration by parts, simplifying leads to an integral of the form \( \int x^{-2} \, dx \).
  • Using the power rule, we find that \( \int x^{-2} \, dx = -x^{-1} + C \), or equivalently \( -\frac{1}{x} + C \).
Understanding and applying the power rule is essential in calculus as it provides a direct and efficient way to tackle integrals with basic power functions, as shown in the step-by-step solution for this exercise.

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Most popular questions from this chapter

Decide whether the statements are true for all continuous functions, \(f\). Give an explanation for your answer. If \(\operatorname{LEFT}(2)<\int_{a}^{b} f(x) d x,\) then \(\operatorname{LEFT}(4)<\) \(\int_{a}^{b} f(x) d x.\)

(a) Find \(\int \sin \theta \cos \theta d \theta\) (b) You probably solved part (a) by making the substitution \(w=\sin \theta\) or \(w=\cos \theta .\) (If not, go back and do it that way.) Now find \(\int \sin \theta \cos \theta d \theta\) by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities $$ \begin{aligned}\sin (2 \theta) &=2 \sin \theta \cos \theta \\\\\cos (2 \theta) &=\cos ^{2} \theta-\sin ^{2} \theta \end{aligned}$$ Find \(\int \sin \theta \cos \theta d \theta\) using one of these identities and then the substitution \(w=2 \theta\) (d) You should now have three different expressions for the indetinite integral \(\int \sin \theta \cos \theta d \theta .\) Are they really different? Are they all correct? Explain.

Explain what is wrong with the statement. $$\int \cos \left(x^{2}\right) d x=\sin \left(x^{2}\right) /(2 x)+C$$.

The rate at which water is flowing into a tank is \(r(t)\) gallons/minute, with \(t\) in minutes. (a) Write an expression approximating the amount of water entering the tank during the interval from time \(t\) to time \(t+\Delta t,\) where \(\Delta t\) is small. (b) Write a Riemann sum approximating the total amount of water entering the tank between \(t=0\) and \(t=5 .\) Write an exact expression for this amount. (c) By how much has the amount of water in the tank changed between \(t=0\) and \(t=5\) if \(r(t)=20 e^{a+2 t} ?\) (d) If \(r(t)\) is as in part (c), and if the tank contains 3000 gallons initially, find a formula for \(Q(t),\) the amount of water in the tank at time \(t\)

Calculate the integrals by partial fractions and then by using the indicated substitution. Show that the results you get are the same.$$\int \frac{2 x}{x^{2}-1} d x ; \text { substitution } w=x^{2}-1$$
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