inttan^(-1)xdx=….+C...

`inttan^(-1)xdx=….+C`

A

`xtan^(-1)x+1/2logabs(1+x^(2))+c`

B

`xtanA^(-1)x-1/2logabs(1+x^(2))+c`

C

`(x-1)tan^(-1)x+c`

D

`xtan^(-1)-logabs(1+x^(2))+c`

Text Solution

Verified by Experts

The correct Answer is:

B

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