`intsin^(-1)((2x)/(1+x^(2)))dx=?`A. `2xtan^(-1)x+log|1+x^(2)|+C`B. `2xtan^(-1)x-log|1+x^(2)|+C`C. `2xsin^(-1)x+log|1+x^(2)|+C`D. none of these

`intsin^(-1)((2x)/(1+x^(2)))dx=?`A. `2xtan^(-1)x+log|1+x^(2)|+C`B. `2x

1.	`intsin^(-1)((2x)/(1+x^(2)))dx=?`A. `2xtan^(-1)x+log\|1+x^(2)\|+C`B. `2xtan^(-1)x-log\|1+x^(2)\|+C`C. `2xsin^(-1)x+log\|1+x^(2)\|+C`D. none of these
Answer» Correct Answer - B Put `x=tant anddx=sec^(2)tdt`. `:." "I=intsin^(-1)((2tant)/(1+tan^(2)t))sec^(2)tdt=sin^(-1)(sin2t)sec^(2)tdt` `=2intunderset(I)(t)underset(II)(sec^(2))tdt`.