3.5 Q-7

Question Statement

Evaluate the integral:
$\int \frac{1}{6x^2 + 5x - 4} , dx$

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Background and Explanation

The given integral involves a rational function where the quadratic expression in the denominator cannot be factored directly into simple terms. To simplify the problem, we use partial fraction decomposition, which breaks the fraction into a sum of simpler terms that can be integrated individually. This technique is especially useful when the denominator factors into linear terms.

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Solution

Step 1: Factorize the Denominator

The denominator $6x^2 + 5x - 4$ can be factored by grouping: $6x^2 + 5x - 4 = 6x^2 + 8x - 3x - 4$
Group and factorize: $(6x^2 + 8x) - (3x + 4) = 2x(3x + 4) - (3x + 4)$
Factor out $(3x + 4)$: $(6x^2 + 5x - 4) = (3x + 4)(2x - 1)$

Thus, the integral becomes: $\int \frac{1}{6x^2 + 5x - 4} , dx = \int \frac{1}{(3x + 4)(2x - 1)} , dx$

Step 2: Partial Fraction Decomposition

Write the fraction as: $\frac{1}{(3x + 4)(2x - 1)} = \frac{A}{3x + 4} + \frac{B}{2x - 1}$

Multiply through by $(3x + 4)(2x - 1)$ to eliminate the denominators: $1 = A(2x - 1) + B(3x + 4)$

Step 3: Solve for A and B

Expand and group terms: $1 = A(2x) - A + B(3x) + B(4)$
$1 = (2A + 3B)x + (-A + 4B)$

Equating coefficients of like terms:

1. Coefficient of $x$: $2A + 3B = 0$
2. Constant term: $-A + 4B = 1$

Solve the system of equations:

• From $2A + 3B = 0$:
  $A = -\frac{3B}{2}$

• Substitute $A = -\frac{3B}{2}$ into $-A + 4B = 1$:
  $-\left(-\frac{3B}{2}\right) + 4B = 1$
  $\frac{3B}{2} + 4B = 1$
  $\frac{11B}{2} = 1 \quad \Rightarrow \quad B = \frac{2}{11}$

• Substitute $B = \frac{2}{11}$ into $A = -\frac{3B}{2}$:
  $A = -\frac{3}{2} \cdot \frac{2}{11} = -\frac{3}{11}$

Step 4: Rewrite the Integral

Substituting $A = -\frac{3}{11}$ and $B = \frac{2}{11}$: $\frac{1}{(3x + 4)(2x - 1)} = \frac{-\frac{3}{11}}{3x + 4} + \frac{\frac{2}{11}}{2x - 1}$

The integral becomes: $\int \frac{1}{6x^2 + 5x - 4} , dx = \int \frac{-\frac{3}{11}}{3x + 4} , dx + \int \frac{\frac{2}{11}}{2x - 1} , dx$

Step 5: Integrate Each Term

1. For the first term: $\int \frac{-\frac{3}{11}}{3x + 4} , dx = -\frac{3}{11} \cdot \frac{1}{3} \ln|3x + 4| = -\frac{1}{11} \ln|3x + 4|$

2. For the second term: $\int \frac{\frac{2}{11}}{2x - 1} , dx = \frac{2}{11} \cdot \frac{1}{2} \ln|2x - 1| = \frac{1}{11} \ln|2x - 1|$

Adding the results: $\int \frac{1}{6x^2 + 5x - 4} , dx = -\frac{1}{11} \ln|3x + 4| + \frac{1}{11} \ln|2x - 1| + C$

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Key Formulas or Methods Used

1. Partial Fraction Decomposition:
   $\frac{1}{(3x + 4)(2x - 1)} = \frac{A}{3x + 4} + \frac{B}{2x - 1}$
2. Logarithmic Integration:
   $\int \frac{1}{ax + b} , dx = \frac{1}{a} \ln|ax + b| + C$
3. Factoring Quadratics:
   $6x^2 + 5x - 4 = (3x + 4)(2x - 1)$

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Summary of Steps

1. Factorize the denominator:
   $6x^2 + 5x - 4 = (3x + 4)(2x - 1)$
2. Apply partial fraction decomposition to rewrite the integrand.
3. Solve for constants $A$ and $B$ using substitution.
4. Rewrite the integral as a sum of simpler fractions.
5. Integrate each term using logarithmic integration rules.

Final Answer:

$\boxed{-\frac{1}{11} \ln|3x + 4| + \frac{1}{11} \ln|2x - 1| + C}$