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:

To integrate $\int \frac{x^4}{x-2} \, dx$, substitute $u = x - 2$, which means $x = u + 2$ and $dx = du$.

1. Expand the numerator:

$x^4 = (u + 2)^4 = u^4 + 4(u^3)(2) + 6(u^2)(4) + 4(u)(8) + 16$

$x^4 = u^4 + 8u^3 + 24u^2 + 32u + 16$

2. Divide by $u$ and integrate:

$\int \frac{u^4 + 8u^3 + 24u^2 + 32u + 16}{u} \, du = \int \left( u^3 + 8u^2 + 24u + 32 + \frac{16}{u} \right) du$

$= \frac{u^4}{4} + \frac{8u^3}{3} + 12u^2 + 32u + 16\log|u| + C_1$

3. Substitute $u = x - 2$ back into the expression:

• $\frac{1}{4}(x-2)^4 = \frac{1}{4}(x^4 - 8x^3 + 24x^2 - 32x + 16)$
• $\frac{8}{3}(x-2)^3 = \frac{8}{3}(x^3 - 6x^2 + 12x - 8) = \frac{8}{3}x^3 - 16x^2 + 32x - \frac{64}{3}$
• $12(x-2)^2 = 12(x^2 - 4x + 4) = 12x^2 - 48x + 48$
• $32(x-2) = 32x - 64$

4. Group the terms by powers of $x$ (absorb constants into $C$):

• Coefficient of $x^4$ ($s$): $\frac{1}{4} = 0.25$
• Coefficient of $x^3$ ($r$): $-2 + \frac{8}{3} = \frac{2}{3} \approx 0.67$
• Coefficient of $x^2$ ($q$): $6 - 16 + 12 = 2$
• Coefficient of $x$ ($p$): $-8 + 32 - 48 + 32 = 8$
• Coefficient of $\log|x-2|$ ($t$): $16$

5. Compare the values:

$t(16) > p(8) > q(2) > r(0.67) > s(0.25)$

Thus, the correct arrangement is $t > p > q > r > s$.