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}$
- DNone of these
✓
Answer
Correct option: A.
$\frac{3}{8}$
a
(a) Putting $x = \tan \theta $, we get
(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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