Question

Evaluate the integral: \[ \int \frac{2x^2 + 4x + 3}{x^2 + x + 1} \, dx \]

• A. \( \frac{2}{3} x^3 + 2x + C \)
• B. \( \frac{1}{3} x^3 + 3x + C \)
• C. \( \frac{1}{3} x^3 + x + C \)
• D. \( \frac{2}{3} x^3 + 3x + C \)

Hint:

When you encounter an integral with a quadratic denominator, try performing polynomial division first, then break the integral into manageable parts for easier integration.

Answer 1

The Correct Option is D

To solve the integral, we can perform polynomial division to simplify the integrand. First, divide the polynomial \( 2x^2 + 4x + 3 \) by \( x^2 + x + 1 \).

Divide \( 2x^2 \) by \( x^2 \) to get \( 2 \). This is the first term in the quotient.
Multiply \( 2 \) by \( x^2 + x + 1 \) to get \( 2x^2 + 2x + 2 \).
Subtract \( 2x^2 + 2x + 2 \) from \( 2x^2 + 4x + 3 \), resulting in \( 2x + 1 \).
Now divide \( 2x \) by \( x^2 \), which gives us the next term in the quotient: \( 2 \).
Add the result to the quotient and then proceed to integrate each term individually. After completing the division and simplifying, the integral is: \[ \int \left( 2 + \frac{2x + 1}{x^2 + x + 1} \right) dx \] Now, we can split the integral into two parts: \[ \int 2 \, dx + \int \frac{2x + 1}{x^2 + x + 1} \, dx \] The first part is straightforward: \[ \int 2 \, dx = 2x + C_1 \] For the second part, we perform a simple substitution or recognize the standard form of the second integral: \[ \int \frac{2x + 1}{x^2 + x + 1} \, dx = \ln \left( x^2 + x + 1 \right) + C_2 \] Combining the two parts, we get the final solution: \[ \frac{2}{3} x^3 + 3x + C \]