MCQ
$\int_0^a {\frac{{x\,dx}}{{\sqrt {{a^2} + {x^2}} }}} = $
  • $a\,(\sqrt 2 - 1)$
  • B
    $a\,(1 - \sqrt 2 )$
  • C
    $a\,(1 + \sqrt 2 )$
  • D
    $2a\sqrt 3 $

Answer

Correct option: A.
$a\,(\sqrt 2 - 1)$
a
(a) Put $t = {a^2} + {x^2} $

$\Rightarrow 2xdx = dt,$ then

$\int_0^a {\frac{{xdx}}{{\sqrt {{a^2} + {x^2}} }} = \frac{1}{2}\int_{{a^2}}^{2{a^2}} {\frac{1}{{\sqrt t }}dt} } $

$ = [{(2{a^2})^{1/2}} - {a^{2/2}}] = a(\sqrt 2 - 1)$.

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