_ ^ 1 + x^2 1 - x^2 dx = - Vidyadip

MCQ

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

✓
$\frac{3}{2}{\sin ^{ - 1}}x - \frac{1}{2}x\sqrt {1 - {x^2}} + c$
B
$\frac{3}{2}{\sin ^{ - 1}}x + \frac{1}{2}x\sqrt {1 - {x^2}} + c$
C
$\frac{3}{2}{\cos ^{ - 1}}x - \frac{1}{2}x\sqrt {1 - {x^2}} + c$
D
$\frac{3}{2}{\cos ^{ - 1}}x + \frac{1}{2}x\sqrt {1 - {x^2}} + c$

✓

Answer

Correct option: A.

$\frac{3}{2}{\sin ^{ - 1}}x - \frac{1}{2}x\sqrt {1 - {x^2}} + c$

a
(a)Put $x = \sin \theta \Rightarrow dx = \cos \theta \,d\theta ,$ then it reduces to $\int_{}^{} {(1 + {{\sin }^2}\theta )\,d\theta } = \theta + \frac{1}{2}\int_{}^{} {(1 - \cos 2\theta )\,d\theta } $
$ = \frac{{3\theta }}{2} - \frac{1}{2}\sin \theta \sqrt {1 - {{\sin }^2}\theta } + c = \frac{3}{2}{\sin ^{ - 1}}x - \frac{1}{2}x\sqrt {1 - {x^2}} + c$.

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