if `I=int(log(t+sqrt(1+t^2)))/sqrt(1+t^2)dt=1/2(g(t))^2+c` then `g(2)` is (A) `2 log(2+sqrt5)` (B) `log (2 + sqrt 5)` (C) `1/sqrt5 log (2 + sqrt5)` (D) `1/2 log ( 2 + sqrt 5)`A. `(1)/(sqrt5)log(2+sqrt5)`B. `(1)/(1)log(2_sqrt5)`C. `2log(2+sqrt5)`D. `log(2+sqrt5)`

if `I=int(log(t+sqrt(1+t^2)))/sqrt(1+t^2)dt=1/2(g(t))^2+c` then `g(2)`

1.	if `I=int(log(t+sqrt(1+t^2)))/sqrt(1+t^2)dt=1/2(g(t))^2+c` then `g(2)` is (A) `2 log(2+sqrt5)` (B) `log (2 + sqrt 5)` (C) `1/sqrt5 log (2 + sqrt5)` (D) `1/2 log ( 2 + sqrt 5)`A. `(1)/(sqrt5)log(2+sqrt5)`B. `(1)/(1)log(2_sqrt5)`C. `2log(2+sqrt5)`D. `log(2+sqrt5)`
Answer» Correct Answer - D `intlog(t+sqrt(1+t^(2)))/(sqrt(1+t^(2)))dt=(1)/(2)(g(t))^(2)+C" …(i)"` `"Let "l=intlog(t+sqrt(1+t^(2)))/(sqrt(1+t^(2)))dt` `"Put "u=log(t+sqrt(1+t^(2)))` `rArr" "du=(1)/(t+sqrt(1+t^(2)))xx(1+(1)/(2sqrt(1+t^(2)))xx2t)dt` `rArr" "du=(1)/(t+sqrt(1+t^(2)))xx(sqrt(1+t^(2))+1)/(sqrt(1+t^(2)))dt` `rArr" "du=(dt)/(sqrt(1+t^(2)))` `therefore" "l=int udu` `rArr" "l=(u^(2))/(2)+C` `rArr" "l=(1)/(2)[log(t+sqrt(1+t^(2)))]^(2)+C" ...(ii)"` From Eps. (i) and (ii), we get `g(t)=log(t+sqrt(1+t^(2)))` `therefore" "g(2)=log(2+sqrt5)`