Answer
\[\frac{-4}{3}\]
Work Step by Step
Let \[I=\int_{0}^{1}\frac{x-1}{\sqrt x}dx\]
Clearly 0 is point of infinite discontinuity of the integrand $\frac{x-1}{\sqrt x}$
So $I$ is improper integral
\[I=\lim_{t\rightarrow 0^+}\int_{t}^{1}\frac{x-1}{\sqrt x}dx\]
\[I=\lim_{t\rightarrow 0^+}\int_{t}^{1}\left(\sqrt x-x^{-\frac{1}{2}}\right)dx\]
\[I=\lim_{t\rightarrow 0^+}\left[\frac{2x^{\frac{3}{2}}}{3}-2x^{\frac{1}{2}}\right]_{t}^{1}\]
\[I=\lim_{t\rightarrow 0^+}\left[\frac{2}{3}-2-\frac{2t^{\frac{3}{2}}}{3}+2t^{\frac{1}{2}}\right]\]
\[I=\frac{2}{3}-2=-\frac{4}{3}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=-\frac{4}{3}$.
