Answer
$$\frac{1}{{\left| u \right|\sqrt {{u^2} - {a^2}} }}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }}} = \frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a} + C \cr
& {\text{Differentiating}} \cr
& \frac{d}{{dx}}\left[ {\frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a} + C} \right] = \frac{1}{a}\left( {\frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a}} \right) + \frac{d}{{dx}}\left[ C \right] \cr
& = \frac{1}{a}\left( {\frac{{\frac{d}{{dx}}\left[ {\frac{{\left| u \right|}}{a}} \right]}}{{\left| u \right|\sqrt {{{\left( {\frac{{\left| u \right|}}{a}} \right)}^2} - 1} }}} \right) + 0 \cr
& = \frac{1}{a}\left( {\frac{{\frac{1}{a}}}{{\left| u \right|\sqrt {\frac{{{u^2}}}{{{a^2}}} - 1} }}} \right) \cr
& {\text{Simplifying}} \cr
& = \frac{1}{a}\left( {\frac{{\frac{1}{a}}}{{\left| u \right|\sqrt {\frac{{{u^2} - {a^2}}}{{{a^2}}}} }}} \right) \cr
& = \frac{1}{a}\left( {\frac{{\frac{1}{{{a^2}}}}}{{\left| u \right|\sqrt {{u^2} - {a^2}} }}} \right) \cr
& = \frac{1}{{\left| u \right|\sqrt {{u^2} - {a^2}} }} \cr} $$
