3.5 Q-8

Question Statement

Evaluate the integral:

$$\int \frac{2x^3 - 3x^2 - x - 7}{2x^2 - 3x - 2} , dx$$

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Background and Explanation

This problem involves integration using polynomial division and partial fraction decomposition. Polynomial division simplifies the numerator by dividing it by the denominator. Once reduced, the fraction is expressed as a sum of simpler terms using partial fractions, which makes the integral easier to solve.

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Solution

Step 1: Simplify the Integral Using Polynomial Division

We start by dividing the numerator $2x^3 - 3x^2 - x - 7$ by the denominator $2x^2 - 3x - 2$. Performing the division:

1. Divide the leading terms: $\frac{2x^3}{2x^2} = x$.
2. Multiply the entire denominator by $x$: $x(2x^2 - 3x - 2) = 2x^3 - 3x^2 - 2x$.
3. Subtract: $(2x^3 - 3x^2 - x - 7) - (2x^3 - 3x^2 - 2x) = x - 7$.

Thus,

$$\frac{2x^3 - 3x^2 - x - 7}{2x^2 - 3x - 2} = x + \frac{x - 7}{2x^2 - 3x - 2}.$$

The integral becomes:

$$\int \frac{2x^3 - 3x^2 - x - 7}{2x^2 - 3x - 2} , dx = \int x , dx + \int \frac{x - 7}{2x^2 - 3x - 2} , dx.$$

Step 2: Solve the First Integral

The first integral is straightforward:

$$\int x , dx = \frac{x^2}{2}.$$

Step 3: Simplify the Second Integral Using Partial Fractions

The remaining term is $\frac{x - 7}{2x^2 - 3x - 2}$. Factor the denominator:

$$2x^2 - 3x - 2 = (x - 2)(2x + 1).$$

Express $\frac{x - 7}{(x - 2)(2x + 1)}$ as:

$$\frac{x - 7}{(x - 2)(2x + 1)} = \frac{A}{x - 2} + \frac{B}{2x + 1}.$$

Step 4: Solve for A and B

Multiply through by the denominator $(x - 2)(2x + 1)$:

$$x - 7 = A(2x + 1) + B(x - 2).$$

Expanding:

$$x - 7 = 2Ax + A + Bx - 2B.$$

Combine like terms:

$$x - 7 = (2A + B)x + (A - 2B).$$

From this, equate coefficients of $x$ and the constant terms:

1. $2A + B = 1$ (coefficient of $x$).
2. $A - 2B = -7$ (constant term).

Solve this system of equations:

1. From $2A + B = 1$, substitute $B = 1 - 2A$ into $A - 2B = -7$: $A - 2(1 - 2A) = -7.$ Simplify: $A - 2 + 4A = -7 \implies 5A = -5 \implies A = -1.$

2. Substitute $A = -1$ into $B = 1 - 2A$: $B = 1 - 2(-1) \implies B = 3.$

Thus:

$$\frac{x - 7}{(x - 2)(2x + 1)} = \frac{-1}{x - 2} + \frac{3}{2x + 1}.$$

Step 5: Solve the Second Integral

The integral becomes:

$$\int \frac{x - 7}{(x - 2)(2x + 1)} , dx = \int \frac{-1}{x - 2} , dx + \int \frac{3}{2x + 1} , dx.$$

1. Solve $\int \frac{-1}{x - 2} , dx$: $\int \frac{-1}{x - 2} , dx = -\ln|x - 2|.$

2. Solve $\int \frac{3}{2x + 1} , dx$: $\int \frac{3}{2x + 1} , dx = \frac{3}{2} \int \frac{1}{2x + 1} , dx = \frac{3}{2} \ln|2x + 1|.$

Step 6: Combine All Results

Adding the results:

$$\int \frac{2x^3 - 3x^2 - x - 7}{2x^2 - 3x - 2} , dx = \frac{x^2}{2} - \ln|x - 2| + \frac{3}{2} \ln|2x + 1| + C.$$

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Key Formulas or Methods Used

1. Polynomial Division: Simplify the rational expression.
2. Partial Fraction Decomposition: Break the fraction into simpler terms.
3. Integration of Logarithmic Functions: $\int \frac{1}{ax + b} , dx = \frac{1}{a} \ln|ax + b| + C.$

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Summary of Steps

1. Perform polynomial division to separate the integral into simpler terms.
2. Factorize the denominator for partial fractions.
3. Solve for constants $A$ and $B$ using substitution and coefficient comparison.
4. Integrate the terms separately, applying logarithmic rules.
5. Combine all terms to write the final solution.