MCQ
$\int {\sqrt {\frac{{1 + x}}{{1 - x}}} \,\,dx = } $
  • A
    $ - {\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$
  • B
    ${\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \, + c$
  • ${\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$
  • D
    $ - {\sin ^{ - 1}}x - \sqrt {{x^2} - 1} \, + c$

Answer

Correct option: C.
${\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$
c
(c)$I = \int {\sqrt {\frac{{1 + x}}{{1 - x}}} dx} $$ = \int {\frac{{1 + x}}{{\sqrt {1 - {x^2}} }}dx} $
$ = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx} } $$ = {\sin ^{ - 1}}x - \sqrt {1 - {x^2}} + c$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free