_ ^ 1 ( x^2 - 1) x^2 + 1 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{({x^2} - 1)\sqrt {{x^2} + 1} }}} \;dx = $

A
$\frac{1}{{2\sqrt 2 }}\log \left\{ {\frac{{\sqrt {1 + {x^2}} + x\sqrt 2 }}{{\sqrt {1 + {x^2}} - x\sqrt 2 }}} \right\} + c$
B
$\frac{1}{{2\sqrt 2 }}\log \left\{ {\frac{{\sqrt {1 + {x^2}} - \sqrt 2 }}{{\sqrt {1 + {x^2}} + \sqrt 2 }}} \right\} + c$
✓
$\frac{1}{{2\sqrt 2 }}\log \left\{ {\frac{{\sqrt {1 + {x^2}} - x\sqrt 2 }}{{\sqrt {1 + {x^2}} + x\sqrt 2 }}} \right\} + c$
D
None of these

✓

Answer

Correct option: C.

$\frac{1}{{2\sqrt 2 }}\log \left\{ {\frac{{\sqrt {1 + {x^2}} - x\sqrt 2 }}{{\sqrt {1 + {x^2}} + x\sqrt 2 }}} \right\} + c$

c
(c) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {\frac{{dx}}{{({x^2} - 1)\sqrt {{x^2} + 1} }} = \int_{}^{} {\frac{{{{\sec }^2}\theta \,d\theta }}{{({{\tan }^2}\theta - 1)\sec \theta }}} } = \int_{}^{} {\frac{{\cos \theta \,d\theta }}{{(2{{\sin }^2}\theta - 1)}}} $
Again put $t = \sin \theta \Rightarrow dt = \cos \theta \,d\theta ,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{(2{t^2} - 1)}}} = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2} - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}} = \frac{1}{{2\sqrt 2 }}\log \left( {\frac{{t - \frac{1}{{\sqrt 2 }}}}{{t + \frac{1}{{\sqrt 2 }}}}} \right) + c$
$ = \frac{1}{{2\sqrt 2 }}\log \left( {\frac{{\sqrt {1 + {x^2}} - x\sqrt 2 }}{{\sqrt {1 + {x^2}} + x\sqrt 2 }}} \right) + c$.

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