inttan^(-1)xdx=….+C...

`inttan^(-1)xdx=….+C`

A

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

B

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

C

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

D

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

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The correct Answer is:

B

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