Answer
$$x = \frac{1}{2} + \frac{1}{2}{\tan ^2}\left( {\frac{{2 + \pi }}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^x {\frac{1}{{t\sqrt {2t - 1} }}} dt = 1 \cr
& {\text{Integrate by the substitution method}}{\text{, let }} \cr
& y = \sqrt {2t - 1} ,\,\,\,\,\,\,t = \frac{{{y^2} + 1}}{2} \to \frac{1}{t} = \frac{2}{{{y^2} + 1}} \cr
& dy = \frac{1}{{\sqrt {2t - 1} }}dt,\,\,\,\,dt = \sqrt {2t - 1} dy \to dt = ydy \cr
& \cr
& {\text{write in terms of }}y \cr
& \int {\frac{1}{{t\sqrt {2t - 1} }}} dt = \int {\frac{2}{{{y^2} + 1}}\left( {\frac{1}{y}} \right)\left( {ydy} \right)} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\int {\frac{1}{{{y^2} + 1}}dy} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\tan ^{ - 1}}y + C \cr
& {\text{write in terms of }}t \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\tan ^{ - 1}}\sqrt {2t - 1} + C \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \left[ {2{{\tan }^{ - 1}}\sqrt {2t - 1} } \right]_1^x = 1 \cr
& 2\left[ {{{\tan }^{ - 1}}\sqrt {2x - 1} - {{\tan }^{ - 1}}\sqrt {2\left( 1 \right) - 1} } \right] = 1 \cr
& {\tan ^{ - 1}}\sqrt {2x - 1} - {\tan ^{ - 1}}\left( 1 \right) = \frac{1}{2} \cr
& {\text{solving for }}x \cr
& {\tan ^{ - 1}}\sqrt {2x - 1} - \frac{\pi }{4} = \frac{1}{2} \cr
& {\tan ^{ - 1}}\sqrt {2x - 1} = \frac{1}{2} + \frac{\pi }{4} \cr
& {\tan ^{ - 1}}\sqrt {2x - 1} = \frac{{2 + \pi }}{4} \cr
& \sqrt {2x - 1} = \tan \left( {\frac{{2 + \pi }}{4}} \right) \cr
& 2x - 1 = {\tan ^2}\left( {\frac{{2 + \pi }}{4}} \right) \cr
& x = \frac{1}{2} + \frac{1}{2}{\tan ^2}\left( {\frac{{2 + \pi }}{4}} \right) \cr} $$
