$\int {\cfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$ equals

Option b is correct

Let $I = \int {\cfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \cfrac{1}{2}\int {\cfrac{{dx}}{{\sqrt { - {x^2} + \cfrac{9}{4}x} }}}$

$= \cfrac{1}{2}\int {\cfrac{{dx}}{{\sqrt { - \left( {{x^2} - \cfrac{9}{4}x} \right)} }} = \cfrac{1}{2}} \int {\cfrac{{dx}}{{\sqrt {{{\left( {\cfrac{9}{8}} \right)}^2} - \left[ {{x^2} - \cfrac{9}{4}x + {{\left( {\cfrac{9}{8}} \right)}^2}} \right]} }}}$

$= \cfrac{1}{2}\int {\cfrac{{dx}}{{\sqrt {{{\left( {\cfrac{9}{8}} \right)}^2} - {{\left( {x - \cfrac{9}{8}} \right)}^2}} }} = \cfrac{1}{2}{{\sin }^{ - 1}}\left( {\cfrac{{x - \cfrac{9}{8}}}{{\cfrac{9}{8}}}} \right)} + C$

$= \cfrac{1}{2}{\sin ^{ - 1}}\left( {\cfrac{{8x - 9}}{9}} \right) + C$

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