Question Statement

Evaluate the integral:

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

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Background and Explanation

This problem involves integrating a rational function where the denominator contains a repeated linear factor $(x+2{)}^2$. When dealing with repeated linear factors in partial fraction decomposition, we must include terms for each power of the repeated factor up to the highest power present.

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Solution

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

To solve this integral, we use the method of partial fraction decomposition. Since the denominator has a repeated linear factor $(x+2{)}^2$, we express the integrand as:

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

Multiplying both sides of equation (1) by $(x+2{)}^2$ to clear the denominators:

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

Finding the constant $B$:

Substitute $x=-2$ into equation (2) to eliminate the term containing $A$:

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

Therefore, $B=1$.

Finding the constant $A$:

From equation (2), expand the right-hand side:

$3x+7=Ax+2A+B$

Comparing the coefficients of $x$ on both sides:

$3=A$

Therefore, $A=3$.

Rewriting the integrand:

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

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

Integrating both sides:

Now we integrate the decomposed form with respect to $x$:

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

Thus, the final solution is:

$I=3\ln (x+2)-\frac{1}{x+2}+c$

(Note: Technically, this should be written as $3\ln |x+2|-\frac{1}{x+2}+c$ to account for the domain where $x<-2$)

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Key Formulas or Methods Used

• Partial Fraction Decomposition for repeated linear factors: $\frac{P(x)}{(x-a{)}^n}=\frac{A_1}{x-a}+\frac{A_2}{(x-a{)}^2}+\cdots +\frac{A_n}{(x-a{)}^n}$
• Substitution method for finding constants in partial fractions (cover-up method for $B$)
• Coefficient comparison method for finding constants in partial fractions (for $A$)
• $\int \frac{1}{x+a}dx=\ln |x+a|+c$
• Power rule for integration: $\int (x+a{)}^{n}dx=\frac{(x+a{)}^{n+1}}{n+1}+c$ for $n\ne -1$

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Summary of Steps

1. Set up partial fractions: Express $\frac{3x+7}{(x+2{)}^2}$ as $\frac{A}{x+2}+\frac{B}{(x+2{)}^2}$ for a repeated linear factor
2. Clear denominators: Multiply through by $(x+2{)}^2$ to get $3x+7=A(x+2)+B$
3. Find $B$: Substitute $x=-2$ (the root) to eliminate $A$, yielding $B=1$
4. Find $A$: Expand and compare coefficients of $x$ to get $A=3$
5. Decompose: Rewrite the original fraction as $\frac{3}{x+2}+\frac{1}{(x+2{)}^2}$
6. Integrate term by term:
   • Integrate $\frac{3}{x+2}$ to get $3\ln (x+2)$
   • Integrate $(x+2{)}^{-2}$ using the power rule to get $-\frac{1}{x+2}$
7. Combine and add constant: Final answer is $3\ln (x+2)-\frac{1}{x+2}+c$