Problem 51 - Thomas Calculus

Question

Evaluate each integral in Exercises \(47-52\) by reducing the improper fraction and using a substitution (if necessary) to reduce it to standard form. $$ \int \frac{4 t^{3}-t^{2}+16 t}{t^{2}+4} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \(2t^2 - t + 8 \ln|t^2 + 4| + C\).

Step 1: Polynomial Long Division

First, notice that the degree of the numerator (3) is higher than the degree of the denominator (2). Perform polynomial long division to simplify the fraction. Divide \(4t^3 - t^2 + 16t\) by \(t^2 + 4\). The first term in the quotient is \(4t\) since \(4t \cdot t^2 = 4t^3\). Subtract \(4t^3\) from the numerator to get \(-t^2 + 16t\) and continue dividing with \(t\cdot t^2\), followed by \(-1\cdot t^2\) yielding a remainder of 16t.

Step 2: Simplify the Integral

After performing the polynomial long division, the integral simplifies to: \[\int \left(4t + \frac{-1\cdot t^2 + 16t}{t^2 + 4}\right) \, dt\]. This can be written as \[\int 4t \, dt + \int \frac{-t^2 + 16t}{t^2 + 4} \, dt\].

Step 3: Integrate Each Part Separately

First, integrate the polynomial part \(4t\): \[\int 4t \, dt = 2t^2 + C_1\]. Next, simplify the fraction part. The fraction \(\frac{-t^2 + 16t}{t^2 + 4}\) can be split as \(\int -1 \, dt + \int \frac{16t}{t^2 + 4} \, dt\). The first part integrates to \(-t + C_2\).

Step 4: Substitution for Remaining Integral

For the term \(\int \frac{16t}{t^2 + 4} \, dt\), use the substitution \(u = t^2 + 4\), so \(du = 2t \, dt\). This means \(8 \, du = 16t \, dt\). The integral becomes \(8 \int \frac{1}{u} \, du\), which results in \(8 \ln|u| + C_3\). Back-substitute \(u = t^2 + 4\) to get \(8 \ln|t^2 + 4|\).

Step 5: Combine and Simplify Results

Combine all parts to write the complete solution: \[2t^2 - t + 8 \ln|t^2 + 4| + C\], where \(C = C_1 + C_2 + C_3\) is the integration constant. This is the evaluated integral of the given expression.

Key Concepts

Improper Fraction

In integral calculus, an improper fraction often appears when the degree of the numerator is equal to or larger than the degree of the denominator. To simplify the integration of such a fraction, transforming it into a proper fraction is crucial.

• Improper fractions in calculus arise mainly in rational functions within integrals.
• The goal is to transform these into simpler expressions for easier integration, commonly through polynomial long division.

Recognizing when a fraction is improper is your first step. Compare the degrees. If they're equal, or the numerator is greater, you're dealing with an improper fraction that needs reworking before proceeding with other integration methods.

Polynomial Long Division

Polynomial long division is a technique used to transform an improper fraction into a more tractable form for integration. This mirrors the long division process with numbers but focuses on polynomial terms. Here's how it works:

• Identify and divide the highest degree term of the numerator by the highest degree term of the denominator. This becomes the first term of your quotient.
• Multiply this term by the entire divisor and subtract from the original numerator.
• Continue this process with the remainder until the degree of the remainder is less than that of the divisor.

The result is a mixture of a polynomial and a proper fraction, facilitating easier integration with the next steps of dividing the problem into smaller parts.

Substitution Method

The substitution method, also known as u-substitution, is a powerful tool for simplifying integrals, especially when dealing with complex functions. This technique works by transforming the integral through a change of variables, making it easier to evaluate.The core elements of the substitution method include:

• Choosing a substitution that simplifies part of the integrand, typically setting a component equal to a new variable, such as setting \(u = t^2 + 4\).
• Finding the differential of your new variable, \(du\), often requiring dividing or multiplying back into the integral to maintain validity.
• Substituting these into the integral to transform it into an easier form, often reducing to a standard integral form.

After integrating, remember to back-substitute the original variable to return to terms of the original integral.

Integration by Parts

Integration by parts is an essential technique for solving integrals of products of functions, typically denoted as \(\int u \, dv = uv - \int v \, du\). This method leverages the product rule for differentiation in reverse.Steps to apply integration by parts:

• Identify two parts of the integrand—one as \(u\) and the other as \(dv\).
• Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\).
• Substitute these into the formula to transform the integral into a simpler one, which may be easier to evaluate.

While integration by parts wasn't directly used in the original solution, it's always crucial to understand as it complements other methods such as substitution, especially when dealing with polynomials multiplied by exponential or trigonometric functions.