$\int {\sqrt {1 + {x^2}} dx}$ is equal to
$\int {\sqrt {1 + {x^2}} dx}$ is equal to
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
Option a is correct
Let $I = \int {\sqrt {1 + {x^2}} dx}$
$= \cfrac{x}{2}\sqrt {1 + {x^2}} + \cfrac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C$
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