Let  ${\mathcal{l}}_1=\underset{x\to 0}{\lim }\frac{1}{x^3}\underset{0}{\overset{x}{\int }}\frac{t(e^{3t}-1)\mathcal{l}n(1+2t)}{(t^3+3)(1-\cos \sqrt{t})}dt$  and ${\mathcal{l}}_2=\underset{x\to 0}{\lim }\frac{1}{x}(\underset{\frac{\pi }{4}}{\overset{\upsilon }{\int }}\mathcal{l}o{g}_{\frac{1}{2}}{\sin }^{2}tdt-\underset{(\frac{\pi }{4}+x)}{\overset{\upsilon }{\int }}\mathcal{l}o{g}_{\frac{1}{2}}{\sin }^{2}tdt)$ , where $\pi <\upsilon <2\pi$ . Then which of the following is(are) CORRECT?

For ${\mathcal{l}}_1$: Using  $L^{/}$ Hospital rule and Leibnitz’s rule, we get
Given limit = $\underset{x\to 0}{\lim }\left(\frac{x(e^{3x}-1)\mathcal{l}n(1+2x)}{3x^2(x^3+3)(1-\cos \sqrt{x})}\right)$
=  $\underset{x\to 0}{\lim }\frac{\left(\frac{e^{3x}-1}{3x}\right)\frac{\mathcal{l}n(1+2x)}{2x}\times 3x\times 2x}{3x(x^3+3)\frac{(1-\cos \sqrt{x})}{{\left(\sqrt{x}\right)}^2}{\left(\sqrt{x}\right)}^2}$
= $\frac{\left(1\right)\left(1\right)\left(2\right)}{\left(3\right)\left(\frac{1}{2}\right)}=\frac{4}{3}$
So,  ${\mathcal{l}}_1=\frac{4}{3}$