Integrals — Class 12 Maths Solution

Let $I = \int {\sqrt {1 + 3x - {x^2}} } dx = \int {\sqrt {1 - ({x^2} - 3x)} } dx$

$= \int {\sqrt {1 - \left( {{x^2} - 3x + \cfrac{9}{4}} \right) + \cfrac{9}{4}} dx} = \int {\sqrt {{{\left( {\cfrac{{\sqrt {13} }}{2}} \right)}^2} - {{\left( {x - \cfrac{3}{2}} \right)}^2}} dx}$

$= \left[ {\cfrac{{x - \cfrac{3}{2}}}{2} \cdot \sqrt {\cfrac{{13}}{4} - {{\left( {x - \cfrac{3}{2}} \right)}^2}} + \cfrac{{13}}{8}{{\sin }^{ - 1}}\left( {\cfrac{{x - \cfrac{3}{2}}}{{\cfrac{{\sqrt {13} }}{2}}}} \right)} \right] + C$

$= \cfrac{{2x - 3}}{4}\sqrt {1 + 3x - {x^2}} + \cfrac{{13}}{8}{\sin ^{ - 1}}\left( {\cfrac{{2x - 3}}{{\sqrt {13} }}} \right) + C$