Answer
$$\sqrt {t - 1} + {\tan ^{ - 1}}\sqrt {t - 1} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {t - 1} }}{{2t}}} dt \cr
& {\text{substitute }}u = \sqrt {t - 1} ,{\text{ }}t = {u^2} + 1,{\text{ }}dt = 2udu \cr
& \int {\frac{{\sqrt {t - 1} }}{{2t}}} dt = \int {\frac{u}{{2\left( {{u^2} + 1} \right)}}} \left( {2udu} \right) \cr
& = \int {\frac{{{u^2}}}{{{u^2} + 1}}} du \cr
& {\text{use long division}} \cr
& = \int {\left( {1 - \frac{1}{{{u^2} + 1}}} \right)} du \cr
& {\text{sum rule}} \cr
& = \int {du} + \int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{find the antiderivatives}} \cr
& = u + {\tan ^{ - 1}}u + C \cr
& {\text{ replacing }}u = \sqrt {t - 1} \cr
& = \sqrt {t - 1} + {\tan ^{ - 1}}\sqrt {t - 1} + C \cr} $$
