Answer
$6$
Work Step by Step
Here, we can see that $f(x)=\dfrac{1}{|\sqrt x|}$ is not defined at $0$. However, $f(x)=\dfrac{1}{|\sqrt x|}=\dfrac{1}{-\sqrt x}$ for all $x \in [-1,0)$ and $f(x)=\dfrac{1}{|\sqrt x|}=\dfrac{1}{\sqrt x}$ for all $x \in (0,4]$
Thus, we have $\int_{-1}^{0} \dfrac{1}{|\sqrt x|}+\int_{0}^{4} \dfrac{1}{|\sqrt x|}=\lim\limits_{p \to 0^{-}} \int_{-1}^{p} \dfrac{1}{|\sqrt x|}+\lim\limits_{q \to 0^{+}} \int_{q}^{4} \dfrac{1}{|\sqrt x|}$
This implies that
$\lim\limits_{p \to 0^{-}} (-2 \sqrt{-p}+2)+\lim\limits_{q \to 0^{+}} (4-2\sqrt q)=2+4=6$
