Applying the substitution theorem in integrals when the substitution function is not $C^1$.

Question

I was given a simple problem of finding $\int_1^2 \frac{\sqrt{t-1}}{t}dt$. Educated in Physics as I am, I jumped right in and substituted $u^2 = t-1$, $2udu =dt$ and simplified the integral to an immediately solvable one. However, now that I have to pay attention to the theorems that I'm actually using, my book says that in order to be able to use the substitution theorem you need in this case $u$ to be continuous and have a continuous derivative, and in this case $u'$ is not continuous (not even defined!) in $1$. My first idea was to use another theorem I have that says that if $f$ is bounded in $[a,b]$ and is integrable in $[c,b]$ for all $c \in (a,b)$, then $f$ is integrable in $[a,b]$ and I can find the integral as a limit, that would allow me to circumvent the point $x=1$, but that seems too complicated for a simple problem and I was wondering if there wasn't an easier way to go about it.

Answer 1

Why not treat it like any other "problem integral" and start with $$ \lim_{L \rightarrow 1^+} \int_L^2 \frac{\sqrt{1-t}}{t} \,\mathrm{d}t $$