If \(\int\displaystyle\frac{x^4}{x-2}\,dx = px + qx^2 + rx^3 + sx^4 + t\log|x - 2| + C\), where \(C\) is an arbitrary constant and \(p, q, r, s, t\) are real numbers, then the correct arrangement of \(p, q, r, s, t\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(t \gt p \gt q \gt r \gt s\)
Since \(x^4 - 16 = (x - 2)(x^3 + 2x^2 + 4x + 8)\), the division of \(x^4\) by \(x - 2\) gives
\(\displaystyle \frac{x^4}{x - 2} = \frac{(x^4 - 16) + 16}{x - 2}\)
\(\displaystyle = x^3 + 2x^2 + 4x + 8 + \frac{16}{x - 2}\)
Integrating term by term:
\(\displaystyle \int \frac{x^4}{x - 2}\,dx = \frac{x^4}{4} + \frac{2x^3}{3} + 2x^2 + 8x + 16\log|x - 2| + C\)
Comparing coefficients with \(px + qx^2 + rx^3 + sx^4 + t\log|x - 2| + C\):
\(\displaystyle p = 8,\quad q = 2,\quad r = \frac{2}{3},\quad s = \frac{1}{4},\quad t = 16\)
So \(t \gt p \gt q \gt r \gt s\).