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Magazine Chapter 7: Problem 62
Calculate. $$\int \frac{d x}{x\left[1+(\ln x)^{2}\right]}$$.
Short Answer
Expert verified
The short answer for the given integral is:
\(\int \frac{1}{x[1+(\ln x)^2]} dx = \arctan(\ln x) + C\).
Step by step solution
01
Identify Suitable Substitution
One potential substitution is setting \(u = \ln x\), so the expression in the denominator becomes simpler and more manageable. Since \(u = \ln x\), we also have \(x = e^u\). Additionally, we need to convert the differential, \(dx\), to its corresponding form in terms of \(du\). To do this, we will differentiate \(u = \ln x\) with respect to \(x\):
$$\frac{du}{dx} = \frac{1}{x}$$
Thus, we can write \(dx = x \,du\).
02
Substitute and Simplify
Now we will substitute \(u = \ln x\), \(x = e^u\), and \(dx = x \,du\) in the integral:
$$\int \frac{1}{x[1+(\ln x)^2]} dx = \int \frac{1}{e^u[1+u^2]} (e^u \, du)$$
The expression now simplifies to:
$$\int \frac{1}{1+u^2} du$$
03
Recognize the Arctangent Function
We recognize the resulting integrand as the derivative of the arctangent function:
$$\frac{d}{du}(\arctan u) = \frac{1}{1 + u^2}$$
So, the integral becomes:
$$\int \frac{1}{1 + u^2} du = \arctan u + C$$
where \(C\) is the constant of integration.
04
Substitute Back
Now we will substitute back the original variable, \(x\), by using the substitution we made earlier, \(u = \ln x\):
$$\arctan u + C = \arctan (\ln x) + C$$
05
State the Final Answer
Our final answer for the given integral is:
$$\int \frac{1}{x[1+(\ln x)^2]} dx = \arctan(\ln x) + C$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
When faced with an integral that is difficult to solve directly, integration by substitution, also known as u-substitution, is a strategy that can simplify the process. The technique involves replacing a part of the integral with a new variable to make the integration more manageable. It's very similar to applying the chain rule in reverse.
For example, in our exercise, the substitution made was to let \( u = \ln x \), transforming the seemingly complex integrand into a much simpler form. Once a suitable substitution is made, the differential \( dx \) must also be expressed in terms of the new variable, \( du \). The key to a smooth substitution is choosing a \( u \) that not only simplifies the integrand but also allows for a straightforward conversion of \( dx \).
After performing the integration with respect to \( u \) and finding the antiderivative, it's critical to substitute back the original variable to complete the solution. This process requires a clear understanding of the relationships between variables and their differentials. In practice, integration by substitution can transform a challenging integral into one for which the antiderivative is readily known or easier to find.
For example, in our exercise, the substitution made was to let \( u = \ln x \), transforming the seemingly complex integrand into a much simpler form. Once a suitable substitution is made, the differential \( dx \) must also be expressed in terms of the new variable, \( du \). The key to a smooth substitution is choosing a \( u \) that not only simplifies the integrand but also allows for a straightforward conversion of \( dx \).
After performing the integration with respect to \( u \) and finding the antiderivative, it's critical to substitute back the original variable to complete the solution. This process requires a clear understanding of the relationships between variables and their differentials. In practice, integration by substitution can transform a challenging integral into one for which the antiderivative is readily known or easier to find.
Natural Logarithm Properties
The natural logarithm function, denoted as \( \ln(x) \), has properties that permit the simplification and reformation of expressions within calculus. One pivotal property is the logarithm of a power, \( \ln(x^a) = a \cdot \ln(x) \), which essentially scales the logarithm by the power. Another fundamental property is that the natural log of the base of natural logarithms, \( e \), is 1: \( \ln(e) = 1 \).
With integrals involving natural logarithms, it's often handy to leverage these properties to manipulate the expression into a more manageable form before proceeding with techniques like integration by substitution. In our specific scenario, knowing that \( u = \ln(x) \), allows us to replace the logarithmic expressions within the integral and, using derivative properties, transform \( dx \) into \( e^u \cdot du \). Such manipulations hinge on a solid grasp of natural logarithm properties, which can dramatically simplify complex integrands.
With integrals involving natural logarithms, it's often handy to leverage these properties to manipulate the expression into a more manageable form before proceeding with techniques like integration by substitution. In our specific scenario, knowing that \( u = \ln(x) \), allows us to replace the logarithmic expressions within the integral and, using derivative properties, transform \( dx \) into \( e^u \cdot du \). Such manipulations hinge on a solid grasp of natural logarithm properties, which can dramatically simplify complex integrands.
Arctangent Function
The arctangent function, commonly written as \( \arctan(x) \), is the inverse of the tangent function within the restricted domain \( -\frac{\pi}{2} < y < \frac{\pi}{2} \). This implies that \( \arctan(x) \) will give an angle \( y \) whose tangent is \( x \) within the specified range.
One of the key points in calculus is that the derivative of the arctangent function with respect to \( x \) is \( \frac{d}{dx}(\arctan(x)) = \frac{1}{1+x^2} \). This derivative allows the arctangent function to emerge naturally when integrating expressions of the form \( \frac{1}{1+u^2} \) 鈥� a notable occurrence in our exercise.
Identifying when an integrand matches the derivative of a fundamental function, such as the arctangent, is a core skill. It simplifies the process of finding indefinite integrals without resorting to more complex techniques. When students encounter an integrand with the structure \( \frac{1}{1+x^2} \) in the future, recognizing it as linked to the arctangent function will aid in prompt and efficient integration.
One of the key points in calculus is that the derivative of the arctangent function with respect to \( x \) is \( \frac{d}{dx}(\arctan(x)) = \frac{1}{1+x^2} \). This derivative allows the arctangent function to emerge naturally when integrating expressions of the form \( \frac{1}{1+u^2} \) 鈥� a notable occurrence in our exercise.
Identifying when an integrand matches the derivative of a fundamental function, such as the arctangent, is a core skill. It simplifies the process of finding indefinite integrals without resorting to more complex techniques. When students encounter an integrand with the structure \( \frac{1}{1+x^2} \) in the future, recognizing it as linked to the arctangent function will aid in prompt and efficient integration.
