To examine the convergence of the improper integrals:
1: $\int_0^{\infty} \frac1{x \log x}dx$
2: $\int_0^{\infty} \frac1{(x +sin^2x) \log x}dx$
In both the cases we see that $0$ and $1$ are point of infinite discontinuity.
Having problem in verifying the convergence. Help lease.
As you pointed out, there is potential trouble at $x=0$ and at $x=1$. There is actually real trouble at these places. However, there is also trouble "at" infinity. We will show that in each case the integral from $2$ to $\infty$ diverges, so each integral diverges.
For $\int_2^\infty \frac{dx}{x\log x}$, calculate $\int_2^b \frac{dx}{x\log x}$. Let $u=\log x$. Then an antiderivative is $\log u$, so our integral is $\log(\log b)-\log(\log 2)$. This blows up (very sedately!) as $b\to\infty$.
For the second integral, note that if $x\ge 2$ then $x+\sin^2 x\le 2x$. So the integral from $2$ to $b$ is $\ge 2(\log(\log b)-\log(\log 2))$. Again, this blows up.
Remark: In fact, the integral from $0$ to $1/2$ diverges, as does the integral from $1/2$ to $1$, as does the integral from $1$ to $2$.