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inttan^(- 1)x dx...

`inttan^(- 1)x dx`

A

`xtan^(-1)x+(1)/(2)log|1+x^(2)|+C`

B

`xtan^(-1)x-(1)/(2)log|1+x^(2)|+C`

C

`-xtan^(-1)x+(1)/(2)log|1+x^(2)|+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Put `tan^(-1)x=t," so taht "x=tant anddx=sec^(2)tdt`.
`:." "I=intunderset(I)(t)underset(II)(sec^(2))tdt`.
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