Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{dx}}{{\sqrt {8 + 3x - {x^2}} }} = } \int {\cfrac{{dx}}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}}$

$= \int {\cfrac{{dx}}{{\sqrt {8 - \left( {{x^2} - 2 \cdot \cfrac{3}{2} \cdot x + \cfrac{9}{4}} \right) + \cfrac{9}{4}} }}} = \int {\cfrac{{dx}}{{\sqrt {\cfrac{{41}}{4} - {{\left( {x - \cfrac{3}{2}} \right)}^2}} }}}$

$= \int {\cfrac{{dx}}{{\sqrt {{{\left( {\cfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \cfrac{3}{2}} \right)}^2}} }}}$

$= {\sin ^{ - 1}}\left( {\cfrac{{x - \cfrac{3}{2}}}{{\cfrac{{\sqrt {41} }}{2}}}} \right) + C = {\sin ^{ - 1}}\left( {\cfrac{{2x - 3}}{{\sqrt {41} }}} \right) + C$