Question

Evaluate the integral: \[ \int \frac{x^4-16x^2+2x+8}{x^3-4x^2+2}dx \]

• A. \( \frac{x^2+8x+c}{2} \)
• B. \( x^2+8x+c \)
• C. \( x^3-4x+c \)
• D. \( \frac{x^2-8x+c}{2} \)

Hint:

For rational expressions involving polynomial division, first divide the highest degree terms, then subtract and continue recursively.

Answer 1

The Correct Option is B

Perform polynomial long division: \[ \frac{x^4 - 16x^2 + 2x + 8}{x^3 - 4x^2 + 2} \] Dividing \( x^4 \) by \( x^3 \) gives \( x \): \[ x(x^3 - 4x^2 + 2) = x^4 - 4x^3 + 2x \] Subtracting: \[ (x^4 - 16x^2 + 2x + 8) - (x^4 - 4x^3 + 2x) = -4x^3 - 16x^2 + 8 \] Continuing division: \[ \frac{-4x^3 - 16x^2 + 8}{x^3 - 4x^2 + 2} \] Dividing \( -4x^3 \) by \( x^3 \) gives \( -4 \): \[ -4(x^3 - 4x^2 + 2) = -4x^3 + 16x^2 - 8 \] Subtracting: \[ (-4x^3 - 16x^2 + 8) - (-4x^3 + 16x^2 - 8) = -8x^2 + 16 \] Resulting integral: \[ x^2 + 8x + c \] Thus, the correct answer is: \[ x^2+8x+c \]