Problem 49 - 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. $$ \int_{\sqrt{2}}^{3} \frac{2 x^{3}}{x^{2}-1} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 7 + \ln 8 \).

Step 1: Long Division of Polynomials

To simplify the integrand \( \frac{2x^3}{x^2-1} \), perform polynomial long division. Divide \( 2x^3 \) by \( x^2 - 1 \). This gives us a quotient of \( 2x \) with a remainder of \( 2x \). So, the fraction can be written as \( 2x + \frac{2x}{x^2-1} \).

Step 2: Integrate the Polynomial Part

The integrable expression is now \( 2x + \frac{2x}{x^2-1} \). First, integrate the polynomial part, \( \int 2x \, dx \). The integral of \( 2x \) is \( x^2 \).

Step 3: Integration by Substitution for Rational Part

For the rational part, \( \int \frac{2x}{x^2-1} \, dx \), use substitution. Let \( u = x^2 - 1 \), thus \( du = 2x \, dx \). The integral becomes \( \int \frac{1}{u} \, du \), which results in \( \ln |u| \).

Step 4: Substitute Back to Original Variable

After integrating, replace \( u \) with \( x^2 - 1 \). Hence, the integral \( \ln |u| \) becomes \( \ln |x^2 - 1| \).

Step 5: Evaluate the Definite Integral

Combine the two integrals: \( x^2 + \ln |x^2 - 1| \). Evaluate from \( \sqrt{2} \) to \( 3 \).- At \( x = 3 \): - Polynomial Part: \( 3^2 = 9 \) - Logarithmic Part: \( \ln |9 - 1| = \ln 8 \)- At \( x = \sqrt{2} \): - Polynomial Part: \( (\sqrt{2})^2 = 2 \) - Logarithmic Part: \( \ln |2 - 1| = \ln 1 = 0 \)Calculate the difference: \((9 + \ln 8) - (2 + 0) = 7 + \ln 8\).

Step 6: Conclusion

The value of the definite integral is \( 7 + \ln 8 \).

Key Concepts

Improper Fractions

An improper fraction is a type of fraction where the numerator is larger than or equal to the denominator. These fractions can sometimes make integration difficult directly, especially when the numerator has a higher degree than the denominator.
To simplify calculations, the fraction can be rewritten using polynomial long division, which separates it into a polynomial expression and a remaining fraction with a lower degree than the denominator.
This simplification gives us two parts to integrate: a polynomial and a simpler rational function, making evaluation easier.

Polynomial Long Division

Polynomial long division is a method used to divide a polynomial by another polynomial of lower or equal degree. This process is similar to long division with numbers.
Here's how it works:

• Divide the leading term of the dividend by the leading term of the divisor to get the first term of the quotient.
• Multiply the entire divisor by this first term and subtract the result from the original dividend.
• Repeat this process with the new polynomial that results from the subtraction.

Through these steps, you break down a complex fraction into simpler parts, which can be individually integrated or further simplified.
In our example, dividing the numerator by the denominator simplifies the expression into a polynomial quotient and a simpler fractional remainder.

Integration by Substitution

Integration by substitution is a powerful technique often used to make integration more manageable. This method is especially useful when dealing with complex rational functions.
The basic idea is to simplify the integral by substituting a part of the integrand with a single variable 'u'. You then transform the entire integrand and expression into the variable 'u', integrate with respect to 'u', and finally substitute back the original expression.
Here's how it works:

• Identify a substitution: Choose a part of the integrand, typically the function inside another function, as 'u'.
• Differentiate your chosen substitution: Calculate 'du' from 'u'.
• Transform the integral: Replace every part of the integrand in 'x' into terms of 'u' and 'du'.
• Integrate: Perform the integration with respect to 'u'.
• Back substitute: Replace 'u' with the original expression in terms of 'x'.

In the given problem, choosing the substitution 'u = x^2 - 1' simplifies the integral significantly to a natural logarithm form, which is easier to solve.

Definite Integrals

A definite integral represents the area under the curve of a function between two specific points on the x-axis. It is calculated by taking the antiderivative, or indefinite integral, of the function and then evaluating it at the upper and lower bounds given.
To compute a definite integral, follow these steps:

• Find the antiderivative (indefinite integral) of the function.
• Evaluate this antiderivative at the upper limit of integration.
• Evaluate the antiderivative at the lower limit of integration.
• Subtract the value at the lower limit from the value at the upper limit to find the area between the curve and the x-axis.

In this exercise, after performing integration steps on both the polynomial and logarithmic parts, the results are evaluated between the given bounds, \( \sqrt{2} \) and \( 3 \), and the difference yields the area under the curve.