For \(x \ne 2\), \(\int\displaystyle\frac{x^4 - 16}{x - 2}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{x^4}{4} + \frac{2x^3}{3} + 2x^2 + 8x + C\)
The numerator is a difference of squares twice over, so it factorises and the denominator cancels.
\(x^4 - 16 = (x^2 - 4)(x^2 + 4)\)
\(x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)\)
Since \(x \ne 2\),
\( \displaystyle \frac{x^4 - 16}{x - 2} = (x + 2)(x^2 + 4) \)
\( = x^3 + 2x^2 + 4x + 8 \)
Integrating term by term with the power rule:
\( \displaystyle \int (x^3 + 2x^2 + 4x + 8)\,dx = \frac{x^4}{4} + \frac{2x^3}{3} + \frac{4x^2}{2} + 8x + C \)
\( \displaystyle = \frac{x^4}{4} + \frac{2x^3}{3} + 2x^2 + 8x + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x^4}{4} + \frac{2x^3}{3} + 4x^2 + 8x + C\) — Incorrect. The integral of \(4x\) is \(2x^2\), not \(4x^2\).
Option (3) \(\displaystyle\frac{x^4}{4} - \frac{2x^3}{3} + 2x^2 - 8x + C\) — Incorrect. This comes from \((x - 2)(x^2 + 4)\), the quotient for division by \(x + 2\), not by \(x - 2\).
Option (4) \(x^3 + 2x^2 + 4x + 8 + C\) — Incorrect. This is only the simplified integrand; it has not been integrated.