Answer
\[4\ln 4-8\]
Work Step by Step
Let \[I=\int_{0}^{4}\frac{\ln x}{\sqrt x}dx\]
Clearly 0 is point of infinite discontinuity of integrand $\large\frac{\ln x}{\sqrt x}$
\[I=\lim_{t\rightarrow 0^+}\int_{t}^{4}\frac{\ln x}{\sqrt x}dx\;\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{\ln x}{\sqrt x}dx\]
Substitute $\;r=\ln x\;\;\;\ldots (2)$
\[\Rightarrow dr=\frac{1}{x}dx\]
\[I_1=\int\frac{x}{\sqrt{x}}\,rdr=\int re^{\frac{r}{2}}dr\]
Using integration by parts
\[I_1=r\int e^{\frac{r}{2}}dr-\int \left((r)'\int e^{\frac{r}{2}}dr\right)dr\]
\[I_1=2re^{\frac{r}{2}}-2\int e^{\frac{r}{2}}dr\]
\[I_1=2re^{\frac{r}{2}}-4e^{\frac{r}{2}}+C\]
Where $C$ is constant of integration
From (2)
\[I_1=2\ln x \: \sqrt{x}-4\sqrt{x}+C\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow 0^+}\left[2\sqrt{x}\ln x-4\sqrt{x}\right]_{t}^{4}\]
\[I=\lim_{t\rightarrow 0^+}\left[4\ln 4-8-2\sqrt{t}\ln t+4\sqrt{t}\right]\]
\[I=4\ln 4-8-2\lim_{t\rightarrow 0^+}\left[\frac{\ln t}{t^{-\frac{1}{2}}}\right] \;\;\;\;\; \left(\frac{\infty}{\infty} form\right)\]
Using L' Hopital's rule
\[I=4\ln 4-8-2\lim_{t\rightarrow 0^+}\left(\frac{\frac{1}{t}}{\frac{-1}{2t^{\frac{3}{2}}}}\right)\]
\[I=4\ln 4-8-2\lim_{t\rightarrow 0^+}\left[-2\sqrt t\right]\]
\[I=4\ln 4-8\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=4\ln 4-8$.
