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

MCQ

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

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

✓

Answer

Correct option: A.

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

a
(a) Put $\log (x + \sqrt {1 + {x^2}} ) = t \Rightarrow \frac{1}{{\sqrt {1 + {x^2}} }}\,dx = dt,$ then

$\int_{}^{} {\frac{{\log (x + \sqrt {1 + {x^2}} )}}{{\sqrt {1 + {x^2}} }}\,dx} = \int_{}^{} {t\,dt} $

$\int_{}^{} {\frac{{{t^2}}}{2}dt} $ $ = \frac{1}{2}{\left[ {\log (x + \sqrt {1 + {x^2}} )} \right]^2} + c$.

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