3.5 Q-1

Question Statement

Evaluate the integral:

$$\int \frac{3x+1}{x^2 - x - 6} , dx$$

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Background and Explanation

This problem involves integration using partial fractions. The integrand’s denominator can be factorized into linear terms, allowing us to break it into simpler fractions. This method requires basic algebraic manipulation to express the integrand as a sum of simpler terms, followed by straightforward integration of each term.

Prerequisite knowledge:

• Factoring quadratic expressions.
• Partial fraction decomposition.
• Basic integration of rational functions of the form $\int \frac{1}{x + c} , dx$.

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Solution

Step 1: Factorize the denominator

The denominator $x^2 - x - 6$ can be factorized as:

$$x^2 - x - 6 = (x + 2)(x - 3)$$

Thus, the integrand becomes:

$$\frac{3x+1}{x^2 - x - 6} = \frac{3x+1}{(x+2)(x-3)}.$$

Step 2: Decompose into partial fractions

We express the fraction as:

$$\frac{3x+1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}.$$

Multiplying through by the denominator $(x+2)(x-3)$ gives:

$$3x + 1 = A(x - 3) + B(x + 2).$$

Step 3: Solve for $A$ and $B$

Expand the right-hand side:

$$3x + 1 = A(x) - 3A + B(x) + 2B = (A + B)x + (-3A + 2B).$$

Equating coefficients of $x$ and the constant term:

1. For $x$: $A + B = 3$.
2. For the constant: $-3A + 2B = 1$.

Solve these equations simultaneously:

• From $A + B = 3$: $B = 3 - A$.
• Substitute $B = 3 - A$ into $-3A + 2B = 1$:

$$-3A + 2(3 - A) = 1 \implies -3A + 6 - 2A = 1 \implies -5A + 6 = 1 \implies A = 1.$$

• Using $A + B = 3$, substitute $A = 1$:

$$1 + B = 3 \implies B = 2.$$

Thus, $A = 1$ and $B = 2$.

Step 4: Rewrite the integral

Using the partial fraction decomposition:

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

The integral becomes:

$$\int \frac{3x+1}{x^2 - x - 6} , dx = \int \frac{1}{x+2} , dx + 2 \int \frac{1}{x-3} , dx.$$

Step 5: Integrate each term

Using the standard formula $\int \frac{1}{x+c} , dx = \ln|x+c| + C$:

1. $\int \frac{1}{x+2} , dx = \ln|x+2| + C_1$.
2. $2 \int \frac{1}{x-3} , dx = 2 \ln|x-3| + C_2$.

Combine the results:

$$\int \frac{3x+1}{x^2 - x - 6} , dx = \ln|x+2| + 2\ln|x-3| + C,$$

where $C = C_1 + C_2$ is the constant of integration.

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Key Formulas or Methods Used

1. Partial Fraction Decomposition:

$$\frac{3x+1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}.$$

2. Integration of Rational Functions:

$$\int \frac{1}{x+c} , dx = \ln|x+c| + C.$$

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Summary of Steps

1. Factorize the denominator: $x^2 - x - 6 = (x+2)(x-3)$.
2. Decompose the fraction into partial fractions: $\frac{3x+1}{(x+2)(x-3)} = \frac{1}{x+2} + \frac{2}{x-3}$.
3. Solve for coefficients $A$ and $B$ using simultaneous equations.
4. Rewrite the integral in terms of simpler fractions.
5. Integrate each term using $\int \frac{1}{x+c} , dx = \ln|x+c| + C$.
6. Combine results to get the final solution:

$$\ln|x+2| + 2\ln|x-3| + C.$$