Integrals — Class 12 Maths Solution

Let $I = \int {\sqrt {1 + \cfrac{{{x^2}}}{9}} dx} = \cfrac{1}{3}\int {\sqrt {9 + {x^2}} dx} = \cfrac{1}{3}\int {\sqrt {{x^2} + {3^2}} dx}$

$= \cfrac{1}{3}\left[ {\cfrac{x}{2}\sqrt {{x^2} + 9} + \cfrac{9}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right|} \right] + C$

$= \cfrac{x}{6}\sqrt {x + 9} + \cfrac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + C$

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