that the integrand is undefined at x = 1 x=1x, equals, 1. On an interval that didn't contain this point, the FTC would tell us that F ( x ) = x + 1 x 1 F (x)= x1 x+1 F, prime, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 1, divided by, x, minus, 1, end fraction. But since the function is undefined at the endpoint x = 1 x=1x, equals, 1 of the interval, the derivative is not defined over the whole interval