3.5 Q-5

Question Statement

Evaluate the integral:
$\int \frac{3 - x}{1 - x - 6x^2} , dx$

---

Background and Explanation

The given problem involves a rational function where the numerator is linear, and the denominator is a quadratic polynomial that factors into two linear terms. The solution involves partial fraction decomposition to simplify the integral into manageable components, followed by the application of logarithmic integration.

---

Solution

Step 1: Factorize the Denominator

The denominator $1 - x - 6x^2$ can be factored as: $(1 - x - 6x^2) = (1 + 2x)(1 - 3x)$

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

Step 2: Partial Fraction Decomposition

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

Multiply through by $(1 + 2x)(1 - 3x)$: $3 - x = A(1 - 3x) + B(1 + 2x)$

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

Equating coefficients of like terms:

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

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

1. From $A + B = 3$:
   $B = 3 - A$

2. Substitute into $-3A + 2B = -1$:
   $-3A + 2(3 - A) = -1$ $-3A + 6 - 2A = -1$ $-5A = -7 \quad \Rightarrow \quad A = \frac{7}{5}$

3. Using $B = 3 - A$:
   $B = 3 - \frac{7}{5} = \frac{15}{5} - \frac{7}{5} = \frac{8}{5}$

Step 4: Rewrite the Integral

Substitute $A = \frac{7}{5}$ and $B = \frac{8}{5}$: $\frac{3 - x}{(1 + 2x)(1 - 3x)} = \frac{\frac{7}{5}}{1 + 2x} + \frac{\frac{8}{5}}{1 - 3x}$

Thus: $\int \frac{3 - x}{1 - x - 6x^2} , dx = \int \frac{\frac{7}{5}}{1 + 2x} , dx + \int \frac{\frac{8}{5}}{1 - 3x} , dx$

Step 5: Integrate Each Term

1. For the first term: $\int \frac{\frac{7}{5}}{1 + 2x} , dx = \frac{7}{5} \int \frac{1}{1 + 2x} , dx$ Using the substitution $u = 1 + 2x \Rightarrow du = 2 , dx$: $\int \frac{1}{1 + 2x} , dx = \frac{1}{2} \ln|1 + 2x|$ Thus: $\int \frac{\frac{7}{5}}{1 + 2x} , dx = \frac{7}{10} \ln|1 + 2x|$

2. For the second term: $\int \frac{\frac{8}{5}}{1 - 3x} , dx = \frac{8}{5} \int \frac{1}{1 - 3x} , dx$ Using the substitution $v = 1 - 3x \Rightarrow dv = -3 , dx$: $\int \frac{1}{1 - 3x} , dx = -\frac{1}{3} \ln|1 - 3x|$ Thus: $\int \frac{\frac{8}{5}}{1 - 3x} , dx = -\frac{8}{15} \ln|1 - 3x|$

Step 6: Combine Results

Adding the two parts: $\int \frac{3 - x}{1 - x - 6x^2} , dx = \frac{7}{10} \ln|1 + 2x| - \frac{8}{15} \ln|1 - 3x| + C$

---

Key Formulas or Methods Used

1. Partial Fraction Decomposition:
   $\frac{P(x)}{Q(x)} = \frac{A}{\text{factor1}} + \frac{B}{\text{factor2}}$

2. Logarithmic Integration:
   $\int \frac{1}{ax + b} , dx = \frac{1}{a} \ln|ax + b|$

3. Substitution Method for Simplifying Integrals.

---

Summary of Steps

1. Factorize the denominator:
   $1 - x - 6x^2 = (1 + 2x)(1 - 3x)$
2. Apply partial fraction decomposition:
   $\frac{3 - x}{1 - x - 6x^2} = \frac{A}{1 + 2x} + \frac{B}{1 - 3x}$
3. Solve for $A$ and $B$:
   $A = \frac{7}{5}, \quad B = \frac{8}{5}$
4. Rewrite the integral:
   $\int \frac{3 - x}{1 - x - 6x^2} , dx = \int \frac{\frac{7}{5}}{1 + 2x} , dx + \int \frac{\frac{8}{5}}{1 - 3x} , dx$
5. Integrate each term using logarithmic integration.
6. Final result:
   $\boxed{\frac{7}{10} \ln|1 + 2x| - \frac{8}{15} \ln|1 - 3x| + C}$