Apostol Analytic Number Theory Ch3 ex1 I have worked through Exercise 1 in Chapter 3. Part (a) requires you to show that
$$\sum\limits_{n \le x} {\frac{{\log n}}{n} = \frac{1}{2}{{\log }^2}x + A + O\left( {\frac{{\log x}}{x}} \right)} $$
The Euler summation formula to apply is
$$\sum\limits_{y < n \le x} {f(n) = \,\,\,\int\limits_y^x {f(t)\,\,{\rm{d}}t} } \,\,\, + \,\,\,\int\limits_y^x {\left( {t - \left[ t \right]} \right)f'(t)\,\,{\rm{d}}t} \,\, + \,\,f(x)\left( {\left[ x \right] - x} \right)\,\, + \,\,f(y)\left( {\left[ y \right] - y} \right)$$
I managed to solve it satisfactorily, but I have also looked at other solutions posted online. In particular, one by a chap called Greg Hirst who has kindly made availabe his solutions to the entire book, which can be accessed here
The extract I am referring to in this post is here: 
His approach was clear enough, except one line where he claims,
$$\left| {\int\limits_x^\infty  {\left( {t - \left[ t \right]} \right)\left( {\frac{{1 - \log t}}{{{t^2}}}} \right)\,\,{\rm{d}}t} } \right|\,\, \le \,\,2\int\limits_x^\infty  {\frac{{\log t}}{{{t^2}}}} \,\,{\rm{d}}t$$
I have constructed my own argument why the "2" can be incorportated as shown, but it seems rather awkward.  Is there anyone who reads this post that can explain why this might be the "go to" approach favoured by some people?
Any insight much appreciated.

Note that for $x\geq e$,
$$\left| {\int_x^\infty  {\left( {t - \left[ t \right]} \right)\left( {\frac{{1 - \log t}}{{{t^2}}}} \right)\,dt} } \right|\leq  {\int\limits_x^\infty  { \underbrace{\left({t - \left[ t \right]}\right)}_{<1} \Big( {\frac{{\overbrace{1}^{\leq \log t} + \log t}}{{{t^2}}}} \Big)\,dt} } 
\leq 2\int_x^\infty  {\frac{{\log t}}{{{t^2}}}} \,dt.$$
A  more natural upper bound would be
$$\int_x^\infty  \frac{1}{{t^2}} \,dt+\int_x^\infty  {\frac{{\log t}}{{{t^2}}}} \,dt=\frac{1}{x}+\int_x^\infty  {\frac{{\log t}}{{{t^2}}}} \,dt.$$
But in both cases, at end, we will arrive to $O\left(\frac{\log(x)}{x}\right)$.
