Question Statement

Evaluate the integral:

$\int \frac{9x^2+3x+29}{(x+1)\left(x^2+4\right)}dx$

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Background and Explanation

This problem requires partial fraction decomposition to break down a complex rational function into simpler fractions that can be integrated individually. You should be familiar with equating coefficients and standard integration formulas involving logarithms and inverse tangent functions.

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Solution

We begin by decomposing the rational function into partial fractions. Since the denominator contains a linear factor $(x+1)$ and an irreducible quadratic factor $(x^2+4)$, we set up the decomposition as:

$\frac{9x^2+3x+29}{(x+1)\left(x^2+4\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}$

Finding the constants $A$, $B$, and $C$:

Multiply both sides of equation (1) by $(x+1)(x^2+4)$ to clear the denominators:

$9x^2+3x+29=A\left(x^2+4\right)+(Bx+C)(x+1)$

Step 1: Find $A$ using the cover-up method

Substitute $x=-1$ into equation (2) to eliminate the term with $B$ and $C$:

$\begin{array}{rl}9(-1{)}^2+3(-1)+29& =A\left((-1{)}^2+4\right)+(B(-1)+C)(-1+1)\\ 9(1)-3+29& =A(1+4)+(-B+C)(0)\\ 9-3+29& =5A\\ 35& =5A\\ A& =7\end{array}$

Step 2: Find $B$ and $C$ by comparing coefficients

Expand the right side of equation (2):

$\begin{array}{rl}9x^2+3x+29& =A{x}^2+4A+B{x}^2+Bx+Cx+C\\ 9x^2+3x+29& =(A+B)x^2+(B+C)x+(4A+C)\end{array}$

Comparing coefficients of $x^2$:

$\begin{array}{rl}9& =A+B\\ 9& =7+B\\ B& =2\end{array}$

Comparing coefficients of $x$:

$\begin{array}{rl}3& =B+C\\ 3& =2+C\\ C& =1\end{array}$

Step 3: Rewrite the integrand

Substituting $A=7$, $B=2$, and $C=1$ back into equation (1):

$\frac{9x^2+3x+29}{(x+1)\left(x^2+4\right)}=\frac{7}{x+1}+\frac{2x+1}{x^2+4}$

Step 4: Integrate term by term

Now we integrate both sides with respect to $x$:

$\int \frac{9x^2+3x+29}{(x+1)\left(x^2+4\right)}dx=7\int \frac{1}{x+1}dx+\int \frac{2x+1}{x^2+4}dx$

Split the second integral:

$=7\ln (x+1)+\int \left(\frac{2x}{x^2+4}+\frac{1}{x^2+4}\right)dx$

$=7\ln (x+1)+\int \frac{2x}{x^2+4}dx+\int \frac{1}{x^2+4}dx$

For the first integral, notice that the numerator is the derivative of the denominator ($\frac{d}{dx}(x^2+4)=2x$), so:

$\int \frac{2x}{x^2+4}dx=\ln \left(x^2+4\right)$

For the second integral, use the standard arctangent formula with $a=2$:

$\int \frac{1}{x^2+4}dx=\int \frac{1}{(x{)}^2+(2{)}^2}dx=\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)$

Final Answer:

Combining all terms and adding the constant of integration:

$=7\ln (x+1)+\ln \left(x^2+4\right)+\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+c$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: $\frac{P(x)}{(x-a)(x^2+bx+c)}=\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}$
• Logarithmic Integration: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+c$ (used for $\frac{2x}{x^2+4}$)
• Arctangent Integration: $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$
• Coefficient Comparison Method: Equating coefficients of like powers of $x$ after expanding

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Summary of Steps

1. Set up partial fractions: Express the integrand as $\frac{A}{x+1}+\frac{Bx+C}{x^2+4}$
2. Find $A$: Substitute $x=-1$ to get $A=7$
3. Find $B$ and $C$: Expand and compare coefficients of $x^2$ and $x$ to get $B=2$ and $C=1$
4. Decompose: Rewrite the integrand as $\frac{7}{x+1}+\frac{2x}{x^2+4}+\frac{1}{x^2+4}$
5. Integrate:
   • $\int \frac{7}{x+1}dx=7\ln (x+1)$
   • $\int \frac{2x}{x^2+4}dx=\ln (x^2+4)$ (recognize derivative of denominator)
   • $\int \frac{1}{x^2+4}dx=\frac{1}{2}{\tan }^{-1}(\frac{x}{2})$ (arctan formula)
6. Combine: Add all results and include constant of integration $c$