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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after simplifying the integral and using substitution, we arrive at this result.

Practice All Chapter MCQs →

\int \sin ^5 x \cos ^4 x d x=

f(x)

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

f\left(\frac{4}{25}\right)

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

{f}\left(\frac{1}{2}\right)=2

f(1)

{C}

\int \frac{x^7}{\left(1-x^2\right)^5} d x

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx