MCQ
$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}} \;dx = $
- ✓$ - \frac{{\sqrt {1 + {x^2}} }}{x} + c$
- B$\frac{{\sqrt {1 + {x^2}} }}{x} + c$
- C$ - \frac{{\sqrt {1 - {x^2}} }}{x} + c$
- D$ - \frac{{\sqrt {{x^2} - 1} }}{x} + c$
✓
Answer
Correct option: A.
$ - \frac{{\sqrt {1 + {x^2}} }}{x} + c$
(a) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}\,dx = \int_{}^{} {\frac{{{{\sec }^2}\theta \,d\theta }}{{{{\tan }^2}\theta \sec \theta }} = \int_{}^{} {{\rm{cosec}}\,\theta \,{\rm{cot}}\theta \,d\theta } } } $
$ = - {\rm{cosec}}\,\theta + {\rm{c}} = \frac{{{\rm{-}}\sqrt {{x^{\rm{2}}} + 1} }}{x} + c.$
$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}\,dx = \int_{}^{} {\frac{{{{\sec }^2}\theta \,d\theta }}{{{{\tan }^2}\theta \sec \theta }} = \int_{}^{} {{\rm{cosec}}\,\theta \,{\rm{cot}}\theta \,d\theta } } } $
$ = - {\rm{cosec}}\,\theta + {\rm{c}} = \frac{{{\rm{-}}\sqrt {{x^{\rm{2}}} + 1} }}{x} + c.$
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