<Dl> <Dd> ∫ 1 x d x = x 0 0 + C (\ displaystyle \ int (\ frac (1) (x)) \, dx = (\ frac (x ^ (0)) (0)) \ + C) </Dd> </Dl> <Dd> ∫ 1 x d x = x 0 0 + C (\ displaystyle \ int (\ frac (1) (x)) \, dx = (\ frac (x ^ (0)) (0)) \ + C) </Dd> <P> Instead the integral is given by: </P> <Dl> <Dd> ∫ 1 a 1 x d x = ln ⁡ a, (\ displaystyle \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx = \ ln a,) </Dd> <Dd> ∫ 1 x d x = ln ⁡ x + C . (\ displaystyle \ int (\ frac (1) (x)) \, dx = \ ln x + C .) </Dd> </Dl>

Is the reciprocal of a one to one function also one to one