Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)` isA. `log(1+sqrt2)`B. `log(1+sqrt2)-(pi)/(4)`C. `log(1+sqrt2)+pi//4`D. None of these

Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)`

1.	Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)` isA. `log(1+sqrt2)`B. `log(1+sqrt2)-(pi)/(4)`C. `log(1+sqrt2)+pi//4`D. None of these
Answer» Correct Answer - B On putting `x=tan theta rArr dx=sec^(2) theta d theta`, we get `f(x)=int(tan^(2) theta.sec^(2)theta)/(sec^(2) theta(1+sec theta))d theta` `=int(sec^(2) theta-1)/(1+sec theta)d theta=int(sec theta-1) d theta` `=log (sec theta tan theta)- theta+C` `rArr" "f(x)=log(sqrt(1+x^(2))+x)-tan^(-1)x+C` At `x=0," "f(0)=log(1+0)-0+C rArr C=0` At `x=1," "f(1)=log(1+sqrt2)-(pi)/(4)`