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

MCQ

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

A
$\sqrt {1 + {x^2}} + {\tan ^{ - 1}}x + c$
B
$\sqrt {1 + {x^2}} - \log \{ x + \sqrt {1 + {x^2}} \} + c$
✓
$\sqrt {1 + {x^2}} + \log \{ x + \sqrt {1 + {x^2}} \} + c$
D
$\sqrt {1 + {x^2}} + \log (\sec x + \tan x) + c$

✓

Answer

Correct option: C.

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

c
(c)$\int_{}^{} {\frac{{x + 1}}{{\sqrt {{x^2} + 1} }}\,dx = \int_{}^{} {\frac{x}{{\sqrt {{x^2} + 1} }}\,dx + \int_{}^{} {\frac{1}{{\sqrt {{x^2} + 1} }}\,dx} } } $
Put ${x^2} + 1 = t \Rightarrow 2x\,dx = dt$, then it reduce to $\frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^{1/2}}}} + \int_{}^{} {\frac{1}{{\sqrt {{x^2} + 1} }}} } \,dx = \frac{1}{2}.2.{t^{1/2}} + \log (x + \sqrt {{x^2} + 1} ) + c$
$ = {({x^2} + 1)^{1/2}} + \log (x + \sqrt {{x^2} + 1} ) + c.$

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