Question

In the following exercises, evaluate the definite integral. $$ \int_{1}^{2} \frac{1+2 x+x^{2}}{3 x+3 x^{2}+x^{3}} d x $$

Short answer

The value of the definite integral is \( \ln\left(\frac{26}{7}\right) \).

Step-by-step solution

Simplify the expression

First, observe the expression \( \frac{1+2x+x^2}{3x+3x^2+x^3} \). Notice that the denominator \( 3x + 3x^2 + x^3 \) can be rewritten as \( x(3 + 3x + x^2) \). This simplification doesn't directly help integration but is useful for factoring or comparing orders.

Perform polynomial long division

To simplify, perform polynomial long division of \( 1+2x+x^2 \) by \( 3x+3x^2+x^3 \). The division doesn't work as typically due to degree, indicating simplification or substitution might help. Here, we try another approach of substitution after failed direct division simplification.

Substitute for easier integration

Substitute \( u = 3x + 3x^2 + x^3 \). Determine \( \frac{du}{dx} = 3 + 6x + 3x^2 \), so \( du = (3 + 6x + 3x^2)dx \). This substitution can make the integral simpler by changing variable of integration.

Change limits of integration

Using the substitution \( u = 3x + 3x^2 + x^3 \), find the new limits: when \( x = 1 \), \( u = 7 \), and when \( x = 2 \), \( u = 26 \). The integral becomes \( \int_{7}^{26} \frac{1 + 2x + x^2}{u} \frac{du}{3 + 6x + 3x^2} \).

Evaluate the integral

The integral simplifies to \( \int \frac{du}{u} \), which evaluates to \( \ln|u| + C \). Using the limits of integration 7 and 26, calculate \( \ln|26| - \ln|7| = \ln\left(\frac{26}{7}\right) \).

Compute the definite integral

Now that we have \( \ln\left(\frac{26}{7}\right) \), this value represents the definite integral from step 5. It is the required evaluation for \( \int_{1}^{2} \frac{1+2x+x^2}{3x+3x^2+x^3} \, dx \).

Key concepts

Integration Techniques

When tackling any integral problem, starting with a variety of integration techniques can lead to improved results. These techniques provide different approaches to handle complex integrands effectively, making integration a more manageable task. In this particular exercise, several methods are considered:

• **Direct Integration:** Initially, we might consider directly integrating the fraction provided. But, due to the complexity of the polynomial expressions involved, this method can be cumbersome.
• **Substitution Method:** Used here as an efficient alternative, where a substitution helps to simplify the complex polynomial into a more recognizable form before integration.
• **Polynomial Long Division:** Useful when the degree of the numerator is greater than or equal to the degree of the denominator, which helps simplify the integrand before integration.

Choosing the right technique greatly depends on the form of the integrand and requires a strategic approach to simplify or manipulate the integral for easy calculation.

Polynomial Long Division

Often employed to break down complex rational expressions, polynomial long division simplifies the integration process by resolving the degrees between numerators and denominators.
The process is analogous to numerical long division and follows these steps:

• Align the polynomial expression according to degree from high to low, ensuring each term is noted.
• Divide the first term of the numerator by the first term of the denominator.
• Multiply the whole divisor by this quotient and subtract from the original numerator.
• Repeat the process until the degree of the remainder is less than the degree of the divisor.

In this exercise, polynomial long division offered limited simplification due to the roles of substitution, but knowing this technique helps with complex rational functions frequently seen in calculus problems.

Substitution Method

The substitution method is a powerful and straightforward technique for transforming the variable of integration, thereby simplifying the function involved.
It works by introducing a substitution variable, typically denoted as \( u \), to replace an inner function or a composite term within the integrand:

• Identify the substitution, which is usually a part of the integral that complicates direct integration.
• Determine \( du \) by differentiating your choice of \( u \) with respect to \( x \) to modify the differential \( dx \) into \( du \).
• Rewrite the limits of integration according to the new variable \( u \).
• Once integrated easy, the result must be transformed back into the original variable using the substitution relation.

This method was key in the exercise, using \( u = 3x + 3x^2 + x^3 \), turning the integral into a simpler form: \( \int \frac{du}{u} \). By adjusting the limits accordingly, the integral could be evaluated with ease, highlighting the method’s usefulness in streamlining cumbersome integrals.

Limits of Integration

Definite integrals are characterized by fixed boundaries or limits of integration that define the range over which the function is evaluated. Once a substitution has been made, it's crucial to adjust these limits according to the new variable.
Steps to handling limits of integration after substitution include:

• Identify the original limits on the \( x \)-axis, and calculate corresponding \( u \) values using the substitution formula.
• Replace original limits with new ones to match the variable of integration post-substitution.
• Evaluate the definite integral using these new limits to ensure the area under the curve is correctly determined based on the transformed variable.

In this exercise, upon substituting \( u = 3x + 3x^2 + x^3 \), the original limits \( x = 1 \) and \( x = 2 \) became \( u = 7 \) and \( u = 26 \), respectively. This allowed the evaluation of the integral in terms of \( u \), ensuring accurate computation of the definite integral value.