Asymptotics of two expressions involving logarithms (As I am new to algorithmic complexity so),

EDIT:



please give solutions for large x (means as x->infinity) !
You can subtitute $y=\ln x$ and apply logarithm laws,

$$\ln\left(\frac{y+1}{y^3}\right)=\ln\left(1+\tfrac1y\right)-2\ln(y)=O(1/y)+O(\ln(y)),$$ 
which is both smaller or contained in $O(y^{\frac12})$



After the Edit: first simplify your expression. Then it reads as

$$x\cdot\frac{0.2}{y}\cdot\frac{y+1}{y^3}$$

which is $O(x\ln(x)^{-3})$ which is contained in $o(x)$.
Use backslash for special functions like $\ln$.

You didn't specify whether $x \to 0$ or $x \to \infty$ or something else, so your question is not precise enough. But either way, the technique is always to look for the most significant term and "pull it out" in some way to get to some form with an asymptotic expansion. For example:

$\ln(x+1) - \ln(x-1) \\
 = \ln(x(1+\frac{1}{x})) - \ln(x(1-\frac{1}{x})) \\
 = ( \ln(x) + \ln(1+\frac{1}{x}) ) - ( \ln(x) + \ln(1-\frac{1}{x}) ) \\
 \approx ( \frac{1}{x} ) - ( - \frac{1}{x} ) \text{ as } x \to \infty \\
 = \frac{2}{x}$

The asymptotic expansion used here is the Taylor series.

[Edit: Now that the question has been clarified..]

Use the same technique, since the outer function is increasing, we need to identify the asymptotically greatest term inside and factor it out:

$\ln(\frac{\ln(x)+1}{\ln(x)^3}) \\
 = \ln(\frac{1}{\ln(x)^2}(...)) \\
 = -\ln(\ln(x)^2) + \ln(...) \\
 \approx -\ln(\ln(x)^2) + ... \text{ as } x \to \infty$

again using the Taylor series.