3.5 Q-2

Question Statement

Evaluate the integral:

$$\int \frac{5x+8}{(x+3)(2x-1)} , dx$$

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Background and Explanation

This problem requires using partial fraction decomposition to simplify the integrand into a sum of simpler fractions. The denominator has two linear factors, $(x+3)$ and $(2x-1)$, making it suitable for partial fraction methods. Each term can then be integrated using the basic rule for logarithmic integrals.

Prerequisite knowledge:

• Factoring polynomials and setting up partial fractions.
• Basic integration of the form $\int \frac{1}{x+c} , dx = \ln|x+c| + C$.

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Solution

Step 1: Express the fraction as partial fractions

Write the given fraction as:

$$\frac{5x+8}{(x+3)(2x-1)} = \frac{A}{x+3} + \frac{B}{2x-1}.$$

Multiply through by the denominator $(x+3)(2x-1)$ to eliminate fractions:

$$5x + 8 = A(2x - 1) + B(x + 3). \tag{1}$$

Step 2: Solve for $A$ and $B$

Case 1: Substitute $x = -3$

Substituting $x = -3$ into equation (1):

$$5(-3) + 8 = A(2(-3) - 1) + B(-3 + 3).$$

Simplify:

$$-15 + 8 = A(-6 - 1) + 0 \implies -7 = -7A \implies A = 1.$$

Case 2: Substitute $x = \frac{1}{2}$

Substituting $x = \frac{1}{2}$ into equation (1):

$$5\left(\frac{1}{2}\right) + 8 = A\left(2\left(\frac{1}{2}\right) - 1\right) + B\left(\frac{1}{2} + 3\right).$$

Simplify:

$$\frac{5}{2} + 8 = 0 + B\left(\frac{7}{2}\right) \implies \frac{21}{2} = \frac{7}{2}B \implies B = 3.$$

Thus, $A = 1$ and $B = 3$.

Step 3: Rewrite the integral

Substitute the partial fraction decomposition:

$$\frac{5x+8}{(x+3)(2x-1)} = \frac{1}{x+3} + \frac{3}{2x-1}.$$

The integral becomes:

$$\int \frac{5x+8}{(x+3)(2x-1)} , dx = \int \frac{1}{x+3} , dx + \int \frac{3}{2x-1} , dx.$$

Step 4: Integrate each term

First term:

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

Second term:

Factor out constants where needed:

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

Let $u = 2x - 1$, so $du = 2 , dx$ or $dx = \frac{1}{2} , du$. Substitute:

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

Replace $u = 2x - 1$:

$$\frac{3}{2} \ln|2x-1| + C_2.$$

Step 5: Combine the results

Add the results from both terms:

$$\int \frac{5x+8}{(x+3)(2x-1)} , dx = \ln|x+3| + \frac{3}{2} \ln|2x-1| + C,$$

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

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Key Formulas or Methods Used

1. Partial Fraction Decomposition:

$$\frac{5x+8}{(x+3)(2x-1)} = \frac{A}{x+3} + \frac{B}{2x-1}.$$

2. Integration of Rational Functions:

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

3. Substitution Method: For $\int \frac{1}{2x-1} , dx$, let $u = 2x-1$, then $dx = \frac{1}{2} , du$.

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Summary of Steps

1. Write the integrand using partial fractions: $\frac{5x+8}{(x+3)(2x-1)} = \frac{1}{x+3} + \frac{3}{2x-1}$.
2. Solve for $A$ and $B$ using substitution in $5x + 8 = A(2x - 1) + B(x + 3)$.
3. Rewrite the integral as two separate terms:

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

4. Integrate each term:
   • $\int \frac{1}{x+3} , dx = \ln|x+3|$
   • $\int \frac{3}{2x-1} , dx = \frac{3}{2} \ln|2x-1|$.
5. Combine results into the final answer:

$$\ln|x+3| + \frac{3}{2} \ln|2x-1| + C.$$