I have tried this question by drawing the graphs of the above two functions and found that both are giving asymptote at $Y=0$. But surprisingly integration of $\ln{(x)}$ from $0$ to $1$ is defined but integration of $\text{pow}(x,\ln{(x)})$ is undefined. What can be the reason for this?
The issue is that $x^{\ln{(x)}}$ has a pole at $x=0$, so the function is not integrable (in the standard sense) over any domain containing a neighborhood of $0$. This basically means the function isn't integrable on $(0,1]$. To clarify, if $\epsilon > 0$, then you can rewrite $\int_{0}^{1}x^{\ln\left(x\right)}dx$ as
$$\int_{0}^{1}x^{\ln\left(x\right)}dx = \lim_{\epsilon\to 0} \int_{\epsilon}^{1}x^{\ln\left(x\right)}dx,$$
which diverges to $\infty$, and you can use the substitution $x=e^u$ to convince yourself the integral equals $\infty$.
But for $\ln{(x)}$, even though it's not defined at $0$, it's integrable on $(0,1]$. For $\epsilon > 0$, we have
$$\int_0^1 \ln{(x)} d x = \lim_{\epsilon\rightarrow 0}\int_{\epsilon}^1 \ln{(x)} d x = \lim_{\epsilon\rightarrow 0}\left[x \ln{(x)} - x\right]_{\epsilon}^1= \lim_{\epsilon\rightarrow 0}(-1-\epsilon\ln{(\epsilon)} +\epsilon ) = -1,$$
where you can prove that $\displaystyle \lim_{\epsilon \to 0} \epsilon \ln{(\epsilon)} = 0$.
