_ ^ 1 + x^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt {1 + {x^2}} \;dx = } $

✓
$\frac{x}{2}\sqrt {1 + {x^2}} + \frac{1}{2}\log (x + \sqrt {1 + {x^2}} ) + c$
B
$\frac{2}{3}{(1 + {x^2})^{3/2}} + c$
C
$\frac{2}{3}x{(1 + {x^2})^{3/2}} + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{x}{2}\sqrt {1 + {x^2}} + \frac{1}{2}\log (x + \sqrt {1 + {x^2}} ) + c$

a
(a)$\int_{}^{} {\sqrt {1 + {x^2}} } dx = \frac{x}{2}\sqrt {{x^2} + 1} + \frac{1}{2}\log (x + \sqrt {{x^2} + 1} ) + c$.

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