Question Statement

Evaluate the integral:

$\int \frac{4x+9}{x^2+x-12}dx$

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Background and Explanation

This problem involves integrating a proper rational function (where the degree of the numerator is less than the denominator). The standard approach is to factor the denominator and apply partial fraction decomposition to split the integrand into simpler rational functions that can be integrated using logarithmic rules.

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Solution

Let $I=\int \frac{4x+9}{x^2+x-12}dx$

Factoring the Denominator

First, factor the quadratic expression in the denominator by splitting the middle term:

$\begin{array}{rl}{x}^2+x-12& =x^2+4x-3x-12\\ & =x(x+4)-3(x+4)\\ & =(x+4)(x-3)\end{array}$

Thus, the integral becomes:

$I=\int \frac{4x+9}{(x+4)(x-3)}dx$

Setting Up Partial Fractions

We express the integrand as the sum of two partial fractions with unknown constants $A$ and $B$:

$\frac{4x+9}{(x+4)(x-3)}=\frac{A}{x+4}+\frac{B}{x-3}$

Multiplying both sides by $(x+4)(x-3)$ to eliminate denominators:

$4x+9=A(x-3)+B(x+4)$

Solving for Constants $A$ and $B$

To find $A$: Substitute $x=-4$ into equation (2) (this eliminates the $B$ term):

$\begin{array}{rl}4(-4)+9& =A(-4-3)+B(-4+4)\\ -16+9& =A(-7)+B(0)\\ -7& =-7A\\ A& =1\end{array}$

To find $B$: Substitute $x=3$ into equation (2) (this eliminates the $A$ term):

$\begin{array}{rl}4(3)+9& =A(3-3)+B(3+4)\\ 12+9& =A(0)+B(7)\\ 21& =7B\\ B& =3\end{array}$

Final Integration

Substituting $A=1$ and $B=3$ back into equation (1):

$\frac{4x+9}{(x+4)(x-3)}=\frac{1}{x+4}+\frac{3}{x-3}$

Integrating both sides with respect to $x$:

$\begin{array}{rl}\int \frac{4x+9}{(x+4)(x-3)}dx& =\int \frac{1}{x+4}dx+3\int \frac{1}{x-3}dx\\ \int \frac{4x+9}{x^2+x-12}dx& =\ln (x+4)+3\ln (x-3)+c\end{array}$

(Note: The answer can also be written as $\ln \left|(x+4)(x-3{)}^3\right|+c$ using logarithm properties.)

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Key Formulas or Methods Used

• Factoring quadratics: Splitting the middle term to find common factors
• 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 solve for unknown constants
• Logarithmic integration: $\int \frac{1}{x-a}dx=\ln |x-a|+C$

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Summary of Steps

1. Factor the denominator: Rewrite $x^2+x-12$ as $(x+4)(x-3)$
2. Set up partial fractions: Express $\frac{4x+9}{(x+4)(x-3)}$ as $\frac{A}{x+4}+\frac{B}{x-3}$
3. Clear denominators: Multiply through by $(x+4)(x-3)$ to obtain $4x+9=A(x-3)+B(x+4)$
4. Solve for $A$: Substitute $x=-4$ to get $A=1$
5. Solve for $B$: Substitute $x=3$ to get $B=3$
6. Decompose the integrand: Rewrite as $\frac{1}{x+4}+\frac{3}{x-3}$
7. Integrate term by term: Apply $\int \frac{1}{u}du=\ln |u|+C$ to each fraction
8. State final answer: $\ln (x+4)+3\ln (x-3)+c$