₀^1 ( 1 x -1 ) d x=

Step-by-step solution

The integral I = ₀^1 ( 1 x - 1 ) , d x is evaluated by simplifying the logarithmic argument to ( 1-x x ) . Applying the substitution property _a^b f(x) , d x = _a^b f(a+b-x) , d x with a=0 , b=1 yields I = ₀^1 ( x 1-x ) , d x . Adding both forms of I gives 2I = ₀^1 [ ( 1-x x ) + ( x 1-x ) ] , d x = ₀^1 (1) , d x . Since (1) = 0 , it follows that 2I = 0 , and thus I = 0 . Final answer: 0