$\smallint \,\,\sqrt {\frac{{x\,\, - \,\,1}}{{x\,\, + \,\,1}}\,\,dx} $ is equal to

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

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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \,\,\sqrt {\frac{{x\,\, - \,\,1}}{{x\,\, + \,\,1}}\,\,dx}

is equal to

View answer and explanation

Correct option: B

\sqrt {{x^2} - \,\,1} - \,\,{\sin ^{ - 1}}\,\,x + c

Option B (

\sqrt {{x^2} - \,1} - \,{\sin ^{ - 1}}\,x + c

) is correct because it is the result of simplifying the integrand and using appropriate integration techniques.