Set of values of $x$ for which $1+\log x<x$

Question

Find the set of values of $x$ for which $$1+\log x<x$$

$$ x>0\\ f(x)=\log x+1-x<0\\ f(1)=0\\ f'(x)=\frac{1}{x}-1\\ x>1\implies f'(x)<0\\ 0<x<1\implies f'(x)>0\\ \implies x\in(1,\infty) $$ But, my reference gives the solution $x\in(0,1)\cup(1,\infty)$, why am I missing the additional domain ?

Answer 1

The conclusion reached in the last line is wrong. You need to find the values of $x$ for which $f(x)<0$, but you found the ones for which $f'(x)<0$ instead. For the answer, note that $x=1$ is the point of global maximum of $f(x)$ and $f(1)=0$. The function is strictly increasing for $x<1$ and strictly decreasing for $x>1$. This means $f(x)<0$ for all $x>0,x\ne1$.

Answer 2

$$\frac1x<1\iff x>1.$$

Hence by integration,

$$x>1\iff\int_1^x\frac{d\color{green}x}{\color{green}x}<\int_1^xd\color{green}x\iff\log x<x-1$$

and

$$x<1\iff\int_x^1\frac{d\color{green}x}{\color{green}x}>\int_x^1d\color{green}x\iff-\log x>1-x.$$