Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} dx = \cfrac{1}{2}\int {\cfrac{{2x + 4}}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \cfrac{1}{2}\int {\cfrac{{2x + 2 + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \cfrac{1}{2}\int {\cfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + \int {\cfrac{{dx}}{{\sqrt {{x^2} + 2x + 3} }}} }$

Let $I = {I_1} + {I_2}$
(say)

……(i)
Where ${I_1} = \cfrac{1}{2}\int {\cfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \cfrac{1}{2}\int {\cfrac{{dt}}{t} = \cfrac{1}{2}\int {{t^{ - 1/2}}dt} = \cfrac{1}{2} \times 2{t^{1/2}}}$

$= \cfrac{1}{2} \times 2\sqrt {{x^2} + 2x + 3} + {C_1} = \sqrt {{x^2} + 2x + 3} + {C_1}$

…(ii)
Also, ${I_2} = \int {\cfrac{{dx}}{{\sqrt {{x^2} + 2x + 3} }} = \int {\cfrac{{dx}}{{\sqrt {{x^2} + 2x + 1 - 1 + 3} }}} }$

$= \int {\cfrac{{dx}}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}} = \log \left| {\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 2} } \right|$

$= \log \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right| + {C_2}$

….(iii)
Hence from (i), (ii) and (iii),

we get
$I = \sqrt {{x^2} + 2x + 3} + \log \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right| + C$

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