The `fx^n` y=f(n)` is the sol of diff `eq^n` `(dy)/(dx)+(xy)/(x^2-1)=(x^4+2x)/sqrt(1-x^2)` in `(-1,1)` satisfying `f(0)=0` then `int_(-sqrt3/2)^(sqrt3/2) f(n)ndx` is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2)) in (-1,1) satisfying f(0)=0. Then int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2)) in (-1,1) satisfying f(0)=0. Then int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2)) in (-1,1) satisfying f(0)=0. Then int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2)))in(-1,1) satisfying f(0)=0. Then int_((sqrt(3))/(2))^((sqrt(3))/(2))f(x)dx is

The function y=f(x) is the solution of the differential equation [dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2] in (-1, 1), satisfying f(0)=0. Then int_[-sqrt3/2]^[sqrt3/2] f(x)dx is (A) pi/3 - sqrt3/2 (B) pi/3 - sqrt3/4 (C) pi/6 - sqrt3/4 (D) pi/6 - sqrt3/2

The function y=f(x) is the solution of the differential equation [dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2] in (-1, 1), satisfying f(0)=0 . Then int_[-sqrt3/2]^[sqrt3/2] f(x)dx is (A) pi/3 - sqrt3/2 (B) pi/3 - sqrt3/4 (C) pi/6 - sqrt3/4 (D) pi/6 - sqrt3/2

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2))) in (-1, 1) satisfying f(0)=0 . Then underset((-sqrt(3))/(2))overset((sqrt(3))/(2))intf(x)dx is

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

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