Problem 19 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{4 x^{3}}{x^{4}+1} d x$$

Step-by-Step Solution

Answer

The integral is \( \ln |x^4 + 1| + C \).

Step 1: Identify the Integral Type

The integral \( \int \frac{4x^3}{x^4 + 1} \, dx \) is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests the method of substitution or partial fraction decomposition might be useful.

Step 2: Perform Substitution

Let's use the substitution method, where we set \( u = x^4 + 1 \). Then, the differential is \( du = 4x^3 \, dx \). This means the integral can be rewritten as \( \int \frac{1}{u} \, du \).

Step 3: Integrate Using Basic Formula

The integral \( \int \frac{1}{u} \, du \) is a standard natural logarithm integral, which gives us \( \ln |u| + C \). Therefore, substituting back for \( u \), we have \( \ln |x^4 + 1| + C \).

Step 4: Verify by Differentiation

To verify, differentiate \( \ln |x^4 + 1| \) with respect to \( x \). Using the chain rule, the derivative \( \frac{d}{dx} [\ln |x^4 + 1|] = \frac{1}{x^4 + 1} \cdot 4x^3 \). This simplifies to \( \frac{4x^3}{x^4 + 1} \), which matches the original integrand.

Key Concepts

Rational Functions

A rational function is expressed as the quotient of two polynomials. For example, in our exercise, we have \( \frac{4x^3}{x^4 + 1} \). A distinguishing aspect of rational functions is their behavior when dealing with division between polynomial expressions. The degree of a polynomial is the highest power of its variable. Here, the degree of the numerator is 3 (because of \( 4x^3 \)), and the degree of the denominator is 4 (since it is \( x^4 + 1 \)). This aspect is crucial because it tells us which integration techniques might be suitable.

• If the degree of the numerator is less than that of the denominator, simplification methods like substitution or partial fraction decomposition are often viable choices.
• For rational functions with higher or equal degrees in the numerator compared to the denominator, polynomial long division might be necessary before applying other techniques.

Understanding these functions sets the stage for choosing appropriate integration techniques. This choice is crucial for transforming the integrand into something easier to handle.

Substitution Method

The substitution method is a powerful tool in integral calculus, especially when dealing with rational functions. It revolves around the idea of simplifying an integral by changing variables. In our example, we used substitution to turn a complex integrand into a simpler one.To apply substitution:

• Identify a portion of the integrand that can be replaced with a single variable, often referred to as \( u \).
• Find the differential \( du \), which corresponds to the derivative of your chosen substitution.
• Rewrite the integral in terms of \( u \), ideally simplifying the expression.

For our exercise, setting \( u = x^4 + 1 \) and converting \( dx \) allowed the integral to become \( \int \frac{1}{u} \, du \), a much simpler form that can be easily integrated. Substitution is all about making the integrand match a standard integral form, easing computation.

Partial Fraction Decomposition

Partial fraction decomposition is another method frequently used with rational functions, particularly useful when the degree of the numerator is less than the degree of the denominator and the polynomial in the denominator can be factored. It helps break down complex fractions into sums or differences of simple fractions, which are easier to integrate. Here's a quick step-by-step on how this method usually works:

• Factor the denominator of the integrand, if possible.
• Write the rational function as a sum of fractions, where each term's denominator is one of the factors from the original denominator.
• Find the constants for each fraction by equating expressions or comparing coefficients.
• Integrate each simple fraction separately.

While partial fraction decomposition wasn't directly used in the original problem, understanding its application is valuable. It provides an alternative strategy for tackling integrals where substitution might not be immediately useful, enhancing your mathematical toolkit.