Problem 34 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each integral. } $$ $$ \int \frac{x^{3}+1}{x^{2}+3} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{x^2}{2} - \frac{3}{2}(x^2 + 3) + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C \).

Step 1: Simplify the Expression

Use polynomial long division to simplify the integrand \( \frac{x^3 + 1}{x^2 + 3} \). Divide \( x^3 + 1 \) by \( x^2 + 3 \) to get the quotient and remainder:1. Divide the leading term \( x^3 \) by \( x^2 \) to get \( x \).2. Multiply \( x \) by \( x^2 + 3 \), giving \( x^3 + 3x \).3. Subtract \( x^3 + 3x \) from \( x^3 + 1 \) leaving \( -3x + 1 \).4. Repeat division with \( -3x + 1 \) leading to a remainder of \( -3x + 1 \).The division gives: \( x + \frac{-3x + 1}{x^2 + 3} \).

Step 2: Integrate the Simplified Expression

Integrate each part separately: 1. Integrate \( x \): \( \int x \, dx = \frac{x^2}{2} \).2. Integrate \( \int \frac{-3x}{x^2 + 3} \, dx \).3. Integrate \( \int \frac{1}{x^2 + 3} \, dx \).For \( \int \frac{-3x}{x^2 + 3} \, dx \), use substitution: let \( u = x^2 + 3 \), \( du = 2x \, dx \), adjust the constant as needed. For \( \int \frac{1}{x^2 + 3} \, dx \), use trigonometric substitution or recognize it as a standard form.

Step 3: Apply Integration Techniques

1. For \( \int \frac{-3x}{x^2 + 3} \, dx \), use substitution: - Let \( u = x^2 + 3 \), then \( du = 2x \, dx \), meaning \( \, dx = \frac{1}{2x} \, du \). - Adjust constant: \( -3 \int \frac{x}{x^2 + 3} \, dx = -\frac{3}{2} \int \, du = -\frac{3}{2} u + C_1 \).2. For \( \int \frac{1}{x^2 + 3} \, dx \), recognize the form of the inverse tangent: - Let \( a^2 = 3 \), we use: \( \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C_2 \).

Step 4: Combine Integrated Parts and Simplify

Combine all parts of the integration result:1. \( \frac{x^2}{2} \) from integrating \( x \).2. From substitution: \( -\frac{3}{2}(x^2 + 3) + C_1 \).3. From inverse tangent integration: \( \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C_2 \).Combine:\[ \int \frac{x^3 + 1}{x^2 + 3} \, dx = \frac{x^2}{2} - \frac{3}{2}(x^2 + 3) + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C \]Simplify and combine constants to form the final expression.

Key Concepts

Polynomial Long Division

Polynomial long division is a method used to divide one polynomial by another, similar to numerical long division. It is particularly helpful when dealing with polynomials of higher degrees in the numerator and denominator.
To perform polynomial long division, you follow these steps:

• Identify the leading terms of both the numerator and the denominator.
• Divide the leading term of the numerator by the leading term of the denominator to obtain the first term of the quotient.
• Multiply the entire divisor by the term obtained and subtract the result from the original polynomial.
• Repeat the process using the new polynomial obtained after subtraction.

In our exercise, we divided the polynomial \(x^3 + 1\) by \(x^2 + 3\). This resulted in a quotient of \(x\) and a remainder of \(-3x + 1\). The division step simplifies the original integral and makes it easier to evaluate further.

Substitution Method

The substitution method in integration is akin to changing variables in equations. This technique simplifies the integral by replacing a part of the integrand with a single variable.
Here’s a step-by-step guide to applying substitution:

• Choose a substitution that simplifies the integrand, typically setting \( u \) equal to a function inside the integral.
• Differentiate your substitution equation to express \( dx \) in terms of \( du \).
• Substitute all occurrences of the old variable with the new variable \( u \). Replace \( dx \) with the expression involving \( du \).
• Integrate with respect to \( u \).

In our case, for the part \( \frac{-3x}{x^2 + 3} \) within the integral, let \( u = x^2 + 3 \), then \( du = 2x \, dx \). This substitution converts the integral into a basic form, simplifying the process of integration.

Inverse Trigonometric Integration

Inverse trigonometric integration involves integrals that can be solved using inverse trigonometric functions, such as \( \tan^{-1}, \sin^{-1}, \) or \( \cos^{-1} \).
In our problem, one part of the integral is of the form \( \int \frac{1}{x^2 + a^2} \, dx \), corresponding to the inverse tangent function.
To recognize and solve it:

• Notice the form: \( \int \frac{1}{x^2 + a^2} \, dx \), which matches the inverse tangent: \( \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \).
• Identify \( a^2 \) in the problem, substitute \( a \) appropriately if needed, and then apply the formula.

With \( a = \sqrt{3} \), the integral \( \int \frac{1}{x^2 + 3} \, dx \) becomes \( \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{x}{\sqrt{3}}\right) + C \). This inverse trigonometric form is essential for handling integrals with expressions like \( x^2 + a^2 \) in the denominator.

Integral Calculus

Integral calculus is a branch of calculus focusing on the concept of integration, which is essentially the reverse of differentiation. Integral calculus allows us to find areas, volumes, central points, and many other useful things.
When approaching problems in integral calculus, we typically seek to find an antiderivative, or the original function from which a given derivative was derived.
The key techniques in integral calculus include:

• Identifying and rewriting the integrand to simplify the problem, such as using polynomial long division to split complex fractions.
• Applying substitution to simplify and rewrite the integral in a more manageable form.
• Recognizing standard integral forms, including those involving inverse trigonometric functions, which have known antiderivatives.

In our exercise, using these techniques allowed us to break down a complex integral into easier parts, solve each part, and then combine them to arrive at the final integral solution. This process is foundational to many areas of mathematics and its application in the physical sciences.