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

$2\,(\sqrt x + \,\,{\cot ^{ - 1}}\sqrt x ) + c$

$2\,(\sqrt x - {\cot ^{ - 1}}\,\, - \,\,\sqrt x ) + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: D

2\,(\sqrt x - {\tan ^{ - 1}}\,\,\sqrt x ) + c

Option D (

2\,(\sqrt x - {\tan ^{ - 1}}\,\,\sqrt x ) + c

) is correct because it is the result of integrating the given expression using substitution and algebraic manipulation.