If $\int\frac{x^4}{x-2}dx= px + qx^2 + rx^3+ sx^4+ t\log|x - 2|+ C$, 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) → $t>p>q>r>s$
Given:
$\int \frac{x^4}{x - 2} \, dx = px + qx^2 + rx^3 + sx^4 + t\log|x - 2| + C$
Use polynomial division:
Divide $x^4$ by $(x - 2)$:
$\begin{align*} &\frac{x^4}{x - 2} = x^3 + 2x^2 + 4x + 8 + \frac{16}{x - 2} \\ \Rightarrow\ &\int \frac{x^4}{x - 2} dx = \int (x^3 + 2x^2 + 4x + 8 + \frac{16}{x - 2}) dx \\ =\ &\frac{x^4}{4} + \frac{2x^3}{3} + 2x^2 + 8x + 16\log|x - 2| + C \end{align*}$
Comparing with:
$px + qx^2 + rx^3 + sx^4 + t\log|x - 2| + C$
Match coefficients:
- $s = \frac{1}{4}$
- $r = \frac{2}{3}$
- $q = 2$
- $p = 8$
- $t = 16$
Descending order of $(p, q, r, s, t)$ is:
${t > p > q > r > s}$