Example:
\[\int^2_1 \frac{1}{\sqrt{x-1}} dx\]
.The integrand is discontinuous at
\[x=1\]
.Substitute
\[u=x-1\]
then \[du=dx, \; x=1 \rightarrow u=0, \; x=2 \rightarrow u=1\]
and the integral becomes \[\int^1_0 \frac{1}{u^{1/2}} du= \int^1_0 u^{-1/2} du=[\frac{u^{1/2}}{1/2}]^1_0= 2 \sqrt{1}-2(0)=2\]