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Question

Question: \[\int_{}^{}{\frac{\tan^{- 1}\sqrt{x}}{(1 + x)\sqrt{x}}dx =}\]...

∫tan⁡−1x(1+x)xdx=\int_{}^{}{\frac{\tan^{- 1}\sqrt{x}}{(1 + x)\sqrt{x}}dx =}

A

tan–1 x + c

B

(tan–1 x)2 + c

C

(tan–1x\sqrt{x})2 + c

D

None of these

Answer

(tan–1x\sqrt{x})2 + c

Explanation

Solution

∫tan⁡−1x(1+x)xdx\int \frac { \tan ^ { - 1 } \sqrt { x } } { ( 1 + x ) \sqrt { x } } d x t = tan–1 x\sqrt { \mathrm { x } }

=

=. 12x\frac { 1 } { 2 \sqrt { x } }

= t2 + c 2dt =