Question Statement

Evaluate the integral: $\int \frac{x^3+4x^2+9x+14}{x^2+4x+3}dx$

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Background and Explanation

These problems involve integrating rational functions using partial fraction decomposition. When the degree of the numerator is less than the denominator (proper fraction), we decompose directly. When the degree of the numerator is greater (improper fraction), we first perform polynomial long division to separate the polynomial part from the proper rational remainder.

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Solution

$\int \frac{x^3+4x^2+9x+14}{x^2+4x+3}dx$

Let $I=\int \frac{x^3+4x^2+9x+14}{x^2+4x+3}\mathrm{d}x$

Since the degree of the numerator (3) is greater than the degree of the denominator (2), this is an improper fraction. We use polynomial long division to convert it to a proper fraction.

Step 1: Polynomial Long Division

$\begin{array}{r}x\\ x^2+4x+3x^3+4x^2+9x+14\\ \underline{x^3+4x^2+3x}\\ 6x+14\end{array}$

This gives: $\frac{x^3+4x^2+9x+14}{x^2+4x+3}=x+\frac{6x+14}{x^2+4x+3}$

Factor the denominator: $x^2+4x+3=(x+1)(x+3)$

So the integral becomes: $\int \frac{x^3+4x^2+9x+14}{x^2+4x+3}dx=\int \left(x+\frac{6x+14}{(x+1)(x+3)}\right)dx$

$=\frac{x^2}{2}+\int \frac{6x+14}{(x+1)(x+3)}dx$

Step 2: Partial Fraction Decomposition

Consider: $\frac{6x+14}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$

Multiplying both sides by $(x+1)(x+3)$: $6x+14=A(x+3)+B(x+1)$

Step 3: Find constants $A$ and $B$

Put $x=-1$ in equation (3): $6(-1)+14=A(-1+3)+B(0)$ $8=2A$ $A=4$

Put $x=-3$ in equation (3): $6(-3)+14=A(0)+B(-3+1)$ $-4=-2B$ $B=2$

Substituting back into equation (2): $\frac{6x+14}{(x+1)(x+3)}=\frac{4}{x+1}+\frac{2}{x+3}$

Step 4: Integration

Integrating both sides with respect to $x$: $\int \frac{6x+14}{(x+1)(x+3)}dx=4\int \frac{1}{x+1}dx+2\int \frac{1}{x+3}dx$

$=4\ln |x+1|+2\ln |x+3|$

Substituting into equation (1): $\int \frac{x^3+4x^2+9x+14}{x^2+4x+3}dx=\frac{x^2}{2}+4\ln |x+1|+2\ln |x+3|+C$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: $\frac{P(x)}{(ax+b)(c{x}^2+dx+e)}=\frac{A}{ax+b}+\frac{Bx+C}{c{x}^2+dx+e}$ for irreducible quadratic factors
• Polynomial Long Division: $\frac{\text{Numerator}}{\text{Denominator}}=\text{Quotient}+\frac{\text{Remainder}}{\text{Denominator}}$ when degree of numerator $\ge$ degree of denominator
• Heaviside Cover-Up Method: Substituting roots of linear factors to find partial fraction constants
• Coefficient Comparison: Equating coefficients of like powers of $x$ to solve for unknown constants
• Standard Integrals: $\int \frac{1}{x+a}dx=\ln |x+a|+C$ and $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$

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Summary of Steps

1. Perform polynomial long division to obtain $x+\frac{6x+14}{x^2+4x+3}$
2. Factor the denominator: $(x+1)(x+3)$
3. Set up partial fractions: $\frac{A}{x+1}+\frac{B}{x+3}$
4. Substitute $x=-1$ to find $A=4$
5. Substitute $x=-3$ to find $B=2$
6. Integrate term by term: $\int xdx+4\int \frac{1}{x+1}dx+2\int \frac{1}{x+3}dx$
7. Combine results: $\frac{x^2}{2}+4\ln |x+1|+2\ln |x+3|+C$