Question

In the following exercises, compute each integral using appropriate substitutions. \(\int \frac{d t}{t\left(1+\ln ^{2} t\right)}\)

Short answer

The integral is \(\arctan(\ln t) + C\).

Step-by-step solution

Identify the Need for Substitution

Observe the integral \(\int \frac{dt}{t(1 + \ln^2 t)}\). Notice that the integral can be simplified using a substitution because the presence of \(\ln t\) in the denominator suggests a potential substitution for \(\ln t\).

Choose an Appropriate Substitution

Let \( u = \ln t \). Consequently, the derivative \( \frac{du}{dt} = \frac{1}{t} \) implies \( dt = t \, du \). Therefore, the integral becomes \( \int \frac{1}{t(1 + u^2)} \, t \, du = \int \frac{du}{1 + u^2} \).

Solve the Substituted Integral

The integral \( \int \frac{du}{1 + u^2} \) is a standard integral that results in \( \arctan u + C \), where \( C \) is the constant of integration.

Back-Substitute for the Original Variable

Recall that we set \( u = \ln t \). Substitute back to get \( \int \frac{dt}{t(1 + \ln^2 t)} = \arctan(\ln t) + C \). This represents the antiderivative in terms of the original variable \( t \).

Key concepts

Integral Substitution

Integration can sometimes be challenging when dealing with complex expressions. Fortunately, integral substitution comes to the rescue by simplifying the process. In this concept, we employ an approach where a part of the integrand is replaced with a simpler variable, often represented as \( u \). This substitution aims to transform the integral into a more manageable form.
For instance, take the integral \( \int \frac{1}{t(1 + \ln^2 t)} dt \). Here, the component \( \ln t \) complicates the integral, suggesting an opportunity for substitution. By letting \( u = \ln t \), we simplify our task. The relationships \( \frac{du}{dt} = \frac{1}{t} \) and \( dt = t \, du \) then allow us to recast the integral as \( \int \frac{du}{1 + u^2} \). This method serves as a popular technique to tackle integrals involving complex logarithmic or exponential functions.
Integral substitution not only facilitates easier integration but also sets the stage for solving the integral completely by transforming the variable back to its original form. This method is invaluable in calculus, aiding in tackling integrals that seem impossible at first glance.

Logarithmic Integration

Logarithmic integration primarily deals with integrals that feature a natural logarithm, \( \ln \), within the integrand. This type of integration often requires the strategic application of algebraic manipulation and sometimes substitution to evaluate.
In our exercise, after substituting \( u = \ln t \), the integral morphs into \( \int \frac{du}{1 + u^2} \). This simplified version constitutes a recognizable form related to logarithmic integrals. The presence of \( u^2 \) in the denominator rings a bell for those familiar with standard integrations involving natural logarithms.
Logarithmic integration stands apart due to its capacity to rewrite integrals in more understandable forms. Once transformed, many integrals associated with logarithms can be resolved using known formulas or theorems in calculus. Thus, recognizing when to use logarithmic integration is a skill worth mastering in order to handle sophisticated integration problems that involve \( \ln \) expressions.

Antiderivative Computation

Computing the antiderivative, also known as finding the indefinite integral, is a fundamental task in calculus. It involves determining a function whose derivative is equal to the integrand. This process is integral (pun intended) in understanding how to reverse differentiation.
In the example exercise, after performing substitution, the integral reduces to \( \int \frac{du}{1 + u^2} \). At this stage, recognizing the resulting function is crucial. The integral \( \int \frac{du}{1 + u^2} \) corresponds to the standard antiderivative formula \( \arctan(u) + C \). Here, \( C \) represents the constant of integration, needed because indefinite integrals present a family of solutions.
Finally, retrieving the original variable completes the process: we back-substitute \( u = \ln t \), yielding \( \arctan(\ln t) + C \). This antiderivative computation illustrates the seamless transition from a complicated integrand to a clean result, showcasing how calculus not only simplifies problems but also provides comprehensive solutions.