Answer
$x^{2i}e^{(3+4i)x}=e^{3x}\cos (2\ln x+4x)+i[e^{3x}\sin (2\ln x+4x)]$
Work Step by Step
We will use the Euler's Formula
\[e^{i\theta}=\cos\theta+i\sin\theta\]
$x^{2i}e^{(3+4i)x}=e^{2i\ln x}e^{3x+4ix}$
$x^{2i}e^{(3+4i)x}=e^{3x}e^{2i\ln x+4ix}$
$x^{2i}e^{(3+4i)x}=e^{3x}e^{i(2\ln x+4x)}$
By using Euler's Formula
$x^{2i}e^{(3+4i)x}=e^{3x}\left[\cos (2\ln x+4x)+i\sin (2\ln x+4x)\right]$
$x^{2i}e^{(3+4i)x}=e^{3x}\cos (2\ln x+4x)+i[e^{3x}\sin (2\ln x+4x)]$
Here,
$u(x)=e^{3x}\cos (2\ln x+4x)$ and $v(x)=e^{3x}\sin (2\ln x+4x)$
Hence $x^{2i}e^{(3+4i)x}=e^{3x}\cos (2\ln x+4x)+i[e^{3x}\sin (2\ln x+4x)]$.
