Can I conclude that $ \int_2^x(t\log t)^{-0.9+0.01}dt \le C x^{0.01} $ with sufficiently large $C>0$? Can I conclude that $$ \int_2^x(t\log t)^{-0.9+0.01}dt \le C x^{0.01} $$ with sufficiently large $C>0$?

No. Observe that $\log t\ge\log2$ for $t\ge2$. Then $$ \int_2^x(t\log t)^{-0.89}\,dt\ge(\log2)^{0.89}\int_2^xt^{-0.89}\,dt=\frac{(\log2)^{0.89}}{0.11}\bigl(x^{0.11}-2^{0.11}\bigr). $$