Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
**Evaluate**

\[ \int \frac{2x + 1}{(4x + 3)(x + 1)} \, dx \]

---

In this integral, the function to be integrated is \( \frac{2x + 1}{(4x + 3)(x + 1)} \). The denominator is a product of two linear factors, \( 4x + 3 \) and \( x + 1 \). This is a rational function, which often requires partial fraction decomposition for integration.

**Partial Fraction Decomposition:**
The expression can be decomposed as:

\[ \frac{2x + 1}{(4x + 3)(x + 1)} = \frac{A}{4x + 3} + \frac{B}{x + 1} \]

To find constants \( A \) and \( B \), multiply through by the common denominator and solve the resulting system of equations.

**Integration:**
Once the partial fractions are determined, integrate each term separately:

\[ \int \frac{A}{4x + 3} \, dx + \int \frac{B}{x + 1} \, dx \]

**Result:**
The antiderivative can be found by integrating each term, typically resulting in a combination of logarithmic expressions.
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Transcribed Image Text:**Evaluate** \[ \int \frac{2x + 1}{(4x + 3)(x + 1)} \, dx \] --- In this integral, the function to be integrated is \( \frac{2x + 1}{(4x + 3)(x + 1)} \). The denominator is a product of two linear factors, \( 4x + 3 \) and \( x + 1 \). This is a rational function, which often requires partial fraction decomposition for integration. **Partial Fraction Decomposition:** The expression can be decomposed as: \[ \frac{2x + 1}{(4x + 3)(x + 1)} = \frac{A}{4x + 3} + \frac{B}{x + 1} \] To find constants \( A \) and \( B \), multiply through by the common denominator and solve the resulting system of equations. **Integration:** Once the partial fractions are determined, integrate each term separately: \[ \int \frac{A}{4x + 3} \, dx + \int \frac{B}{x + 1} \, dx \] **Result:** The antiderivative can be found by integrating each term, typically resulting in a combination of logarithmic expressions.
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