Question Statement

Evaluate the integral:

$\int \frac{21-8x}{x^2+x-6}dx$

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Background and Explanation

This problem involves integrating a rational function where the degree of the numerator is less than the degree of the denominator. The key technique is partial fraction decomposition, which breaks the complex fraction into simpler terms of the form $\frac{A}{x-a}$ that integrate directly to logarithmic functions.

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Solution

Let $I$ denote the integral:

$I=\int \frac{21-8x}{x^2+x-6}dx$

Step 1: Factor the denominator

Factor the quadratic expression in the denominator by splitting the middle term:

$\begin{array}{rl}{x}^2+x-6& =x^2+3x-2x-6\\ & =x(x+3)-2(x+3)\\ & =(x+3)(x-2)\end{array}$

Thus, the integral becomes:

$I=\int \frac{21-8x}{(x+3)(x-2)}dx$

Step 2: Set up partial fraction decomposition

Express the rational function as the sum of two simpler fractions:

$\frac{21-8x}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}$

Multiply both sides by $(x+3)(x-2)$ to eliminate the denominators:

$21-8x=A(x-2)+B(x+3)$

Step 3: Solve for the constants $A$ and $B$

To find $A$, substitute $x=-3$ (which makes the $B$ term equal to zero):

$\begin{array}{rl}21-8(-3)& =A(-3-2)+B(-3+3)\\ 21+24& =A(-5)+B(0)\\ 45& =-5A\\ A& =-9\end{array}$

To find $B$, substitute $x=2$ (which makes the $A$ term equal to zero):

$\begin{array}{rl}21-8(2)& =A(2-2)+B(2+3)\\ 21-16& =A(0)+B(5)\\ 5& =5B\\ B& =1\end{array}$

Step 4: Rewrite and evaluate the integral

Substitute $A=-9$ and $B=1$ back into the partial fraction form:

$\frac{21-8x}{(x+3)(x-2)}=\frac{-9}{x+3}+\frac{1}{x-2}$

Now integrate term by term:

$\begin{array}{rl}I& =\int \left(\frac{-9}{x+3}+\frac{1}{x-2}\right)dx\\ & =-9\int \frac{1}{x+3}dx+\int \frac{1}{x-2}dx\\ & =-9\ln (x+3)+\ln (x-2)+c\end{array}$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: For distinct linear factors, $\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$
• Heaviside Cover-up Method: Substituting strategic values of $x$ to eliminate variables and solve for constants
• Logarithmic Integration: $\int \frac{1}{x-a}dx=\ln |x-a|+C$ (presented here as $\ln (x-a)$ per the source)

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Summary of Steps

1. Factor the denominator: Decompose $x^2+x-6$ into $(x+3)(x-2)$
2. Set up partial fractions: Write $\frac{21-8x}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}$
3. Clear denominators: Multiply through by $(x+3)(x-2)$ to obtain $21-8x=A(x-2)+B(x+3)$
4. Find $A$: Substitute $x=-3$ to get $A=-9$
5. Find $B$: Substitute $x=2$ to get $B=1$
6. Separate the integral: Rewrite as $-9\int \frac{1}{x+3}dx+\int \frac{1}{x-2}dx$
7. Integrate: Apply logarithmic integration to obtain $-9\ln (x+3)+\ln (x-2)+c$