dx ( a^2 + x^2 ) ^ 3/2 is equal to - Vidyadip

MCQ

$\int {\frac{{dx}}{{{{({a^2} + {x^2})}^{3/2}}}}} $ is equal to

A
$\frac{x}{{{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}}$
✓
$\frac{x}{{{a^2}{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}}$
C
$\frac{1}{{{a^2}{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}}$
D
None of these

✓

Answer

Correct option: B.

$\frac{x}{{{a^2}{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}}$

b
(b) $I = \int{dx\over{(a^2 + x^2)^3/2}}$
Put $x = a\tan \theta \,\, \Rightarrow dx = a{\sec ^2}\theta \,d\theta $
$\therefore \,I = \int {\frac{{a{{\sec }^2}\theta }}{{{{({a^2} + {a^2}{{\tan }^2}\theta )}^{3/2}}}}d\theta } = \int {\frac{{a{{\sec }^2}\theta }}{{{a^3}{{({{\sec }^2}\theta )}^{3/2}}}}d\theta } $
==> $I = \frac{1}{{{a^2}}}\int {\frac{{d\theta }}{{\sec \theta }}} $ $ = \frac{1}{{{a^2}}}\int {\cos \theta \,d\theta = \frac{1}{{{a^2}}}\sin \theta + c} $
==> $I = \frac{x}{{{a^2}{{({x^2} + {a^2})}^{1/2}}}} + c$.

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