MCQ
$\int_0^\infty {\frac{{dx}}{{{{\left( {x + \sqrt {{x^2} + 1} } \right)}^3}}}} = $
  • $\frac{3}{8}$
  • B
    $\frac{1}{8}$
  • C
    $ - \frac{3}{8}$
  • D
    None of these

Answer

Correct option: A.
$\frac{3}{8}$
a
(a) Putting $x = \tan \theta $, we get

$\int_0^\infty {\frac{{dx}}{{{{\left( {x + \sqrt {{x^2} + 1} } \right)}^3}}}} $

$ = \int_0^{\pi /2} {\frac{{{{\sec }^2}\theta \,d\theta }}{{{{(\tan \theta + \sec \theta )}^3}}}} $

$= \int_0^{\pi /2} {\frac{{\cos \theta }}{{{{(1 + \sin \theta )}^3}}}d\theta } $

$ = \left[ { - \frac{1}{{2{{(1 + \sin \theta )}^2}}}} \right]_0^{\pi /2} $

$= - \frac{1}{8} + \frac{1}{2} = \frac{3}{8}$.

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