Question Statement

Evaluate the integral:

$\int \frac{3x+7}{(x+2)(x+3)}dx$

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Background and Explanation

This problem involves integrating a proper rational function where the degree of the numerator (1) is less than the degree of the denominator (2). We use partial fraction decomposition to break the complex fraction into simpler terms that can be integrated using standard logarithmic rules.

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Solution

Let $I=\int \frac{3x+7}{(x+2)(x+3)}dx$.

We begin by expressing the integrand as a sum of partial fractions. Since the denominator contains two distinct linear factors $(x+2)$ and $(x+3)$, we write:

$\frac{3x+7}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$

To determine the constants $A$ and $B$, we multiply both sides of equation (1) by the common denominator $(x+2)(x+3)$:

$3x+7=A(x+3)+B(x+2)$

Finding the constant $A$:
We substitute $x=-2$ into equation (2). This choice eliminates the $B$ term because $(-2+2)=0$:

$\begin{array}{rl}3(-2)+7& =A(-2+3)+B(-2+2)\\ -6+7& =A(1)+B(0)\\ 1& =A+0\\ A& =1\end{array}$

Finding the constant $B$:
We substitute $x=-3$ into equation (2). This choice eliminates the $A$ term because $(-3+3)=0$:

$\begin{array}{rl}3(-3)+7& =A(-3+3)+B(-3+2)\\ -9+7& =A(0)+B(-1)\\ -2& =0+B(-1)\\ 2& =B\end{array}$

Rewriting the integrand:
Substituting $A=1$ and $B=2$ back into equation (1):

$\frac{3x+7}{(x+2)(x+3)}=\frac{1}{x+2}+\frac{2}{x+3}$

Performing the integration:
Now we integrate both sides with respect to $x$:

$\begin{array}{rl}\int \frac{3x+7}{(x+2)(x+3)}dx& =\int \frac{1}{x+2}dx+2\int \frac{1}{x+3}dx\\ & =\ln (x+2)+2\ln (x+3)+c\end{array}$

Therefore, the final answer is:

$\ln (x+2)+2\ln (x+3)+c$

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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 the roots of the denominator ($x=-2$ and $x=-3$) to isolate and solve for constants $A$ and $B$
• Standard Integral: $\int \frac{1}{x+a}dx=\ln (x+a)+C$

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Summary of Steps

1. Set up partial fraction decomposition: $\frac{3x+7}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$
2. Clear denominators to obtain: $3x+7=A(x+3)+B(x+2)$
3. Substitute $x=-2$ to find $A=1$
4. Substitute $x=-3$ to find $B=2$
5. Rewrite the integral as: $\int \frac{1}{x+2}dx+2\int \frac{1}{x+3}dx$
6. Integrate each term to obtain the final result: $\ln (x+2)+2\ln (x+3)+c$