Problem 78 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{t^{3}-2 t^{2}+3 t-4}{t^{2}+1} d t$$

Step-by-Step Solution

Answer

\( \int \frac{t^{3}-2 t^{2}+3 t-4}{t^{2}+1} dt = \frac{t^2}{2} - 2t + \frac{3}{2} \ln|t^2+1| - 2\tan^{-1}(t) + C \).

Step 1: Identify the Type of Integral

The given integral is \( \int \frac{t^{3}-2t^{2}+3t-4}{t^{2}+1} \, dt \). This is a rational function and we will use polynomial long division as the degree of the numerator is higher than the denominator.

Step 2: Perform Polynomial Long Division

Divide \( t^3 - 2t^2 + 3t - 4 \) by \( t^2 + 1 \). This yields a quotient of \( t - 2 \) and a remainder of \( 3t - 2 \). This allows us to rewrite the integral as \( \int (t - 2) \, dt + \int \frac{3t-2}{t^{2}+1} \, dt \).

Step 3: Integrate the Polynomial Part

The first part of the integral, \( \int (t - 2) \, dt \), is straightforward. We integrate each term to get \( \frac{t^2}{2} - 2t + C_1 \).

Step 4: Integrate the Rational Function Part

For \( \int \frac{3t-2}{t^2+1} \, dt \), separate into two integrals: \( 3 \int \frac{t}{t^2+1} \, dt \) and \(-2 \int \frac{1}{t^2+1} \, dt \).

Step 5: Tackle \( 3 \int \frac{t}{t^2+1} \, dt \)

Use substitution: let \( u = t^2+1 \), then \( du = 2t \, dt \). Thus, \( \int \frac{t}{t^2+1} \, dt = \frac{1}{2} \int \frac{1}{u} \, du \), which is \( \frac{1}{2} \ln|u| + C_2 = \frac{1}{2} \ln|t^2+1| + C_2 \). Multiplying by 3, we have \( \frac{3}{2} \ln|t^2+1| \).

Step 6: Tackle \( -2 \int \frac{1}{t^2+1} \, dt \)

The integral \( \int \frac{1}{t^2+1} \, dt \) is a standard result, equal to \( \tan^{-1}(t) \). Thus, \(-2 \int \frac{1}{t^2+1} \, dt = -2 \tan^{-1}(t) \).

Step 7: Combine All Parts

Combine the results from all previous steps: \( \frac{t^2}{2} - 2t + \frac{3}{2} \ln|t^2+1| - 2\tan^{-1}(t) + C \), where \( C \) is the constant of integration.

Key Concepts

Polynomial Long Division

Polynomial long division is a useful technique when simplifying integrals involving rational functions. A rational function is a division of two polynomials, where the degree (highest power of the variable) of the numerator can determine the method of integration used.

In the given problem, the polynomial in the numerator (\( t^3 - 2t^2 + 3t - 4 \)) has a degree (3) greater than that of the polynomial in the denominator (\( t^2 + 1 \), degree 2). In such cases, polynomial long division simplifies the expression so you can deal with simpler integrals.

Here's what you do:

• Divide the first term of the numerator by the first term of the denominator to get the first term of the quotient, which is \( t \) in this case.
• Multiply the entire divisor by \( t \) and subtract from the numerator to find the new dividend.
• Repeat this process until the degree of the remainder is less than the denominator.

This process results in a quotient of \( t - 2 \) and a remainder of \( 3t - 2 \). You can now rewrite the integral in a more manageable form: \( \int (t - 2) \, dt + \int \frac{3t - 2}{t^2 + 1} \, dt \).

Substitution Method

The substitution method is vital for solving integrals, particularly those that have complex fractions or quotients. In the previous example, dealing with \( \int \frac{3t - 2}{t^2 + 1} \, dt \) involved using substitution.

Here is how you use the substitution method:

• Identify a part of the integral which can be substituted with a single variable, making integration easier.
• For this integral, set \( u = t^2 + 1 \), implying \( du = 2t \, dt \). Since the numerator has a \( t \), this choice of \( u \) makes the fraction cleaner.
• Rewrite the integral in terms of \( u \): \( \int \frac{t}{t^2 + 1} \, dt = \frac{1}{2} \int \frac{1}{u} \, du \).

This process transforms our integral into a natural logarithm function, \( \frac{1}{2} \ln|u| \), and upon converting back to \( t \), you get \( \frac{1}{2} \ln|t^2 + 1| \).

Inverse Trigonometric Functions

Inverse trigonometric functions often appear naturally in integration problems, especially with certain patterns in the integrand.

In our example, \( \int \frac{1}{t^2 + 1} \, dt \) shows up during the division and reduction process. This is a known form that directly integrates to an inverse trigonometric function, specifically, \( \tan^{-1}(t) \).

The integral of \( \int \frac{1}{1 + t^2} \, dt \) has a direct result because of its resemblance to the derivative of \( \tan^{-1}(t) \). Always remembering this form helps with rapid problem-solving when evaluating integrals.

Integration of Rational Functions

Integrating rational functions can often be reduced to simpler steps involving division, substitution, and sometimes recognizing target forms.

When faced with a rational function \( \int \frac{P(t)}{Q(t)} \, dt \), where \( P(t) \) and \( Q(t) \) are polynomials:

• First check if polynomial long division can be applied. If the polynomial in the numerator has a higher degree, start there.
• Rewrite the function as an integral of a polynomial and a proper rational function, as done earlier with \( \int (t - 2) \, dt + \int \frac{3t - 2}{t^2 + 1} \, dt \).
• For the remaining proper rational integral, consider substitution or familiar integral forms like inverse trigonometric functions.

By systematically breaking the integral into known patterns or using substitutions, you simplify the integration process, ultimately arriving at the complete integration: \[ \frac{t^2}{2} - 2t + \frac{3}{2} \ln|t^2+1| - 2\tan^{-1}(t) + C \].