Integrals — Class 12 Maths Solution

Let $\int {\sqrt {{x^2} + 4x - 5} } dx = \int {\sqrt {({x^2} + 4x + 4) - 9} } dx$

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

$= \cfrac{{x + 2}}{2}\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( 3 \right)}^2}} - \cfrac{9}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{{\left( {x + 2} \right)}^2} - {{\left( 3 \right)}^2}} } \right| + C$

$= \cfrac{{x + 2}}{2}\sqrt {{x^2} + 4x - 5} - \cfrac{9}{2}\log \left| {x + 2 + \sqrt {{x^2} + 4x - 5} } \right| + C$

$= \cfrac{{x + 2}}{2}\sqrt {1 - 4x - {x^2}} + \cfrac{5}{2}{\sin ^{ - 1}}\left( {\cfrac{{x + 2}}{{\sqrt 5 }}} \right) + C$