Problem 426 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \( \arctan(\ln t) + C \).

Step 1: Identify the substitution

We begin by analyzing the integral \( \int \frac{dt}{t(1+\ln^2 t)} \). Notice that the function inside the integral suggests that a substitution involving \( \ln t \) might simplify the integrand. Let us consider the substitution \( u = \ln t \), which implies that \( du = \frac{1}{t} dt \). This will help us simplify the integral.

Step 2: Apply the substitution

Using the substitution from Step 1, convert the integral to terms of \( u \). Substituting \( u = \ln t \) and \( du = \frac{1}{t} dt \), the integral becomes: \[ \int \frac{dt}{t(1+\ln^2 t)} = \int \frac{du}{1+u^2} \] This transformation simplifies the expression significantly.

Step 3: Integrate using standard integral

The integral \( \int \frac{du}{1+u^2} \) is a standard form that results in the inverse tangent function. Thus, \[ \int \frac{du}{1+u^2} = \arctan(u) + C \] where \( C \) is the integration constant.

Step 4: Back-substitute in terms of original variable

Now, replace \( u \) with \( \ln t \) to express the result in terms of the original variable \( t \). Therefore, substitute back to get: \[ \arctan(u) + C = \arctan(\ln t) + C \] This is the antiderivative of the original integral.

Step 5: Final solution

The integral of \( \int \frac{dt}{t(1+\ln^2 t)} \) simplifies to: \[ \arctan(\ln t) + C \] This expression is in its simplest form, where \( C \) is the constant of integration.

Key Concepts

Substitution Method

In the realm of integral calculus, the substitution method is a valuable tool for simplifying complex integrals. It allows us to transform a complicated function into a simpler one that is easier to integrate. By using substitution, we aim to replace variables such that the integral becomes easier to solve.
Typically, this involves choosing a new variable for substitution that can simplify the integrand's expression. You may determine the substitution by looking for a common function derivative or a nested function within the integrand.

• First, decide on a substitution. In this problem, the substitution is: let \( u = \ln t \).
• Next, express \( du \) in terms of \( dt \). Here \( du = \frac{1}{t} dt \).
• Then, rewrite the integral in terms of \( u \) and \( du \).

This technique effectively reduces the integral to a simpler form, which can be recognized as a standard integral.

Standard Integral

Once you have transformed the integral using the substitution method, the next step is to identify if it fits any standard integral. Standard integrals are common integral forms that have known solutions, eliminating the need for further complex manipulation.
In our exercise, the integral simplifies to \( \int \frac{du}{1+u^2} \).

• This is a standard integral because it directly matches the form \( \int \frac{1}{1+u^2} \, du \).
• The antiderivative of this standard form is the inverse tangent function, represented as \( \arctan(u) + C \).

Using standard integral forms speeds up the process of finding antiderivatives and makes solving otherwise complex integrals much more manageable.

Inverse Trigonometric Functions

Inverse trigonometric functions appear frequently in calculus, especially when dealing with certain types of standard integrals. Each has a specific form and a corresponding integration result.
In our specific example, the standard integral \( \int \frac{du}{1+u^2} \) yields the inverse tangent function \( \arctan(u) + C \).

• The inverse tangent function, \( \arctan \), reverses the \( \tan \) function, providing the angle whose tangent is the given number.
• When integrating, it often helps in scenarios where you encounter quotients that resemble a Pythagorean identity.

Remember to substitute back into the original variable after finding the antiderivative, as we did with \( u = \ln t \), resulting in \( \arctan(\ln t) + C \). This ensures your solution reflects the proper function in terms of the original variable used before substitution.