Answer
The given improper integral
$$
\int_{0}^{4} \frac{\ln x}{\sqrt{x}} d x=4 \ln 4-8.
$$
Work Step by Step
$$
\int_{0}^{4} \frac{\ln x}{\sqrt{x}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{\ln x}{\sqrt{x}} $ has the vertical asymptote $x =0 $. Since the infinite discontinuity occurs at the left endpoint of $ [0, 4] $ , we must use part (b) of Definition 3 :
$$
\begin{aligned} \int_{0}^{4} \frac{\ln x}{\sqrt{x}} d x &= \lim _{t \rightarrow 0^{+}} \int_{t}^{4} \frac{\ln x}{\sqrt{x}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=\ln x, \quad\quad dv= \frac{dx}{ \sqrt x } }\\ {d u= \frac{dx}{x }, \quad\quad v= 2\sqrt x }\end{array}\right] , \text { then }\\
&=\lim _{t \rightarrow 0^{+}} \left[ 2\sqrt x \ln x- 2\int_{t}^{4} \frac{dx}{ \sqrt x } \right]\\
& = \lim _{t \rightarrow 0^{+}} \left[ 2\sqrt x \ln x- 4 \sqrt x \right]_t^{4}
\\
& =\lim _{t \rightarrow 0^{+}}[(2 \cdot 2 \ln 4-4 \cdot 2)-(2 \sqrt{t} \ln t-4 \sqrt{t})] \\
& =[(2 \cdot 2 \ln 4-4 \cdot 2)-(\lim _{t \rightarrow 0^{+}}(2 \sqrt{t} \ln t) -0)] \\
&=(4 \ln 4-8)-(0-0) \\
& =4 \ln 4-8.
\end{aligned}
$$
where $$ \lim _{t \rightarrow 0^{+}}(2 \sqrt{t} \ln t)=0 $$
by using L'Hospital's rule:
$$
\lim _{t \rightarrow 0^{+}}(2 \sqrt{t} \ln t)=\lim _{t \rightarrow 0^{+}} \frac{2 \ln t}{t^{-1 / 2}} \underset{t \rightarrow 0^{+}}{\lim } \frac{2 / t}{-\frac{1}{2} t^{-3 / 2}}=\lim _{t \rightarrow 0^{+}}(-4 \sqrt{t})=0
$$
Thus the given improper integral is:
$$
\int_{0}^{4} \frac{\ln x}{\sqrt{x}} d x=4 \ln 4-8.
$$
