Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = \int {\cfrac{{x + 2}}{{\sqrt {4 - \left( {{x^2} - 4x + 4} \right)} }}dx}$

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

$= {I_1} + 4{\sin ^{ - 1}}\left( {\cfrac{{x - 2}}{2}} \right) + {C_2}$

…(i)
Where ${I_1} = \int {\cfrac{{\left( {x - 2} \right)dx}}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}}$

Put ${\left( {x - 2} \right)^2} = t$ $\Rightarrow$ $2\left( {x - 2} \right)dx = dt$

$\therefore$ ${I_1} = \cfrac{1}{2}\int {\cfrac{{dt}}{{\sqrt {4 - t} }}}$

$= \cfrac{1}{2}\left[ {\cfrac{{{{\left( {4 - t} \right)}^{ - 1/2 + 1}}}}{{ - \left( {\cfrac{{ - 1}}{2} + 1} \right)}}} \right] + {C_1} = - \sqrt {\left( {4 - t} \right)} + {C_1}$

$= - \sqrt {4 - {{\left( {x - 2} \right)}^2}} + {C_1}$

$= - \sqrt {4 - {x^2} - 4 + 4x} + {C_1} = - \sqrt {4x - {x^2}} + {C_1}$

….(ii)
From (i) and (ii)

we get
$I = - \sqrt {4x - {x^2}} + 4{\sin ^{ - 1}}\left( {\cfrac{{x - 2}}{2}} \right) + C$

$\left[ {C = {C_1} + {C_2}} \right]$