Problem 13 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{8 x-7}{2 x^{2}-x-3} d x$$

Step-by-Step Solution

Answer

The value of the integral is \(2*ln|2x+1| - 3*ln|x-3| + C\).

Step 1: Apply partial fraction decomposition

Rewrite \(\frac{8x-7}{2x^2-x-3}\) in the form of \(A/x-a\) + \(B/x-b\), where A and B are coefficients and a and b are roots of the quadratic equation \(2x^2-x-3 = 0\). Factor the denominator to \(2x^2 - x - 3 = (2x+1)(x-3)\) and set different fractions equal to \(A=(2x+1)\) and \(B=(x-3)\). Use linear system to solve for A and B.

Step 2: Solve the system of equations

Equating the coefficients on each side to form a system of linear equations, we get \(A - 3A = 8\) and \(-B - AB = -7\). Solving this system gives \(A = 2\) and \(B = -3\).

Step 3: Substitute A and B back to the fraction

Substitute A and B back into the fractions \(\frac{8x-7}{2x^2-x-3}\) = \(\frac{2}{2x+1}\) - \(\frac{3}{x-3}\).

Step 4: Perform the Integral

Now, integrate term by term to get \(\int \frac{2}{2x+1} dx - \int \frac{3}{x-3} dx\). This leads to the result \(2*ln|2x+1| - 3*ln|x-3|\).

Step 5: Write the final result

The final answer is \(2*ln|2x+1| - 3*ln|x-3| + C\), where C is the constant of integration. Here, the |x| denotes the absolute value of x.

Key Concepts

Partial Fraction Decomposition

The technique of partial fraction decomposition is valuable for breaking down complex rational expressions into simpler pieces. When you encounter an integral like \( \int \frac{8x-7}{2x^2-x-3} \, dx \), where the denominator is a polynomial, a direct approach to integration isn't straightforward. That's where partial fractions come into play.Here's how it works:

• First, identify the factors of the denominator. In this example, the denominator \(2x^2-x-3\) factors as \((2x+1)(x-3)\).


• After factorization, express the original fraction as a sum of simpler fractions. Specifically, set \(\frac{8x-7}{2x^2-x-3} = \frac{A}{2x+1} + \frac{B}{x-3}\), where \(A\) and \(B\) are constants you'll solve for.

This decomposition simplifies the integration process by turning a tricky rational function into a sum of fractions that are easier to integrate.

Linear System of Equations

After expressing the fraction as a sum of simpler fractions, we need to determine the unknown coefficients \(A\) and \(B\). To find these, substitute back into the equation:\[ 8x-7 = A(x-3) + B(2x+1) \]Expanding and equating coefficients gives us a linear system of equations:

• For the coefficient of \(x\): \(A \, \cdot \, 1 + B \, \cdot \, 2 = 8\)


• For the constant term: \(-3A + B = -7\)

Solving this system, we find \(A = 2\) and \(B = -3\). This step requires careful algebraic manipulation but transforms our original problem into something manageable.

Factorization

Factorization is a crucial step in simplifying and solving expressions and equations, particularly in partial fractions. The denominator of our fraction, \(2x^2-x-3\), needs to be expressed as a product of simpler expressions.For quadratic expressions like this one, set it equal to zero to find the roots through factorization:\[ 2x^2-x-3 = (2x+1)(x-3) \]This process of finding factors involves determining the values where each factor equals zero. Thus, solving gives you the simplified algebraic forms \(2x+1=0\) and \(x-3=0\). This step ensures you can rewrite the original integral in terms of simpler, easy-to-handle fractions.

Integration of Rational Functions

Once the rational function is decomposed into partial fractions, the integration itself becomes straightforward. Each simpler fraction can be integrated individually:- For the integral \(\int \frac{2}{2x+1} \, dx\), use a simple \(\ln\) rule: \( 2 \cdot \ln|2x+1| \).
- For \(\int \frac{-3}{x-3} \, dx\), it simplifies similarly: \(-3 \cdot \ln|x-3|\).The linear property of integration allows you to sum these results together. Therefore, the integral of the original expression becomes the combination of these terms: \(2 \cdot \ln|2x+1| - 3 \cdot \ln|x-3| + C\). This incremental and logical approach helps in breaking down and solving complex integrals efficiently.