_ 1 2 ^ 2 ( e^x-1 e^x+1 ) d x=

Step-by-step solution

The integral I = _ 1 2 ^ 2 ( e^x-1 e^x+1 ) dx has limits 1 2 = - 2 and 2 , forming the symmetric interval [- 2, 2] . Consider the function f(x) = ( e^x-1 e^x+1 ) . Substituting -x gives f(-x) = ( e^ -x -1 e^ -x +1 ) . Multiplying numerator and denominator by e^x yields f(-x) = ( 1-e^x 1+e^x ) = (- e^x-1 e^x+1 ) = - ( e^x-1 e^x+1 ) = -f(x) . Since f(x) is an odd function and the interval is symmetric about zero, the definite integral vanishes: I = 0 . Final Answer: 0