Answer
$x^{2-i}=x^2\cos(\ln x)+i[-x^2\sin (\ln x)]$
Work Step by Step
We will use the Euler's formula
\[e^{i\theta}=\cos \theta+i\sin\theta\]
$x^{2-i}=x^2x^{-i}$
$x^{2-i}=x^2e^{-i\ln x}$
$x^{2-i}=x^2e^{i(-\ln x)}$
By using Euler's Formula
$x^{2-i}=x^2\left[\cos(-\ln x)+i\sin (-\ln x)\right]$
$x^{2-i}=x^2\left[\cos(\ln x)-i\sin (\ln x)\right]$
$x^{2-i}=x^2\cos(\ln x)+i[-x^2\sin (\ln x)]$
Here $\;u(x)=x^2\cos(\ln x)$ and $v(x)=-x^2\sin (\ln x)$
Hence $\;\;x^{2-i}=x^2\cos(\ln x)+i[-x^2\sin (\ln x)]$.
