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In this question, the OP asked to prove that $-int_x^infty fraccos(t)t,dt=gamma+log(x)+int_0^x fraccos(t)-1t,dt$, where $gamma $ is the Euler-Mascheroni constant. However, the two posted answers are incomplete and seem unsatisfactory.

Using Complex Anaysis

One can show for $x>0$ that $-int_x^infty fraccos(t)t,dt=gamma+log(x)+int_0^x fraccos(t)-1t,dt$ using contour integration. To wit, Cauchy's Integral Theorem guarantees that

$$beginalign
0&=oint_C frace^izz,dz\\
&=int_epsilon^R frace^ixx,dx +int_0^pi/2frace^iRe^iphiRe^iphi,iRe^iphi,dphi+int_R^epsilon frace^-xix,i,dx+int_pi/2^0 frace^iepsilon e^iphiepsilon e^iphi,iepsilon e^iphi,dphi\\
&=int_epsilon^R frace^ixx,dx -int_epsilon^R frace^-xx,dx-ifracpi2+O(epsilon)+Oleft(frac1Rright)tag1
endalign$$

whence after taking the real part of both sides of $(1$ and integrating by parts the integral $int_epsilon^R frace^-xx,dx$ with $u=e^-x$ and $v=log(x)$, we find

$$beginalign-int_x^R fraccos(x')x',dx'&=log(x)-int_epsilon^R e^-xlog(x),dx+int_epsilon^xfraccos(x')-1x',dx'\\
&-log(epsilon) -e^-Rlog(R)+e^-epsilonlog(epsilon)+Oleft(frac1Rright)+O(epsilon)tag2
endalign$$

Letting $Rtoinfty$ and $epsilonto 0$ in $(2)$ yields the sought relationship.

Using Real Analysis
Alternatively, we can use either the Laplace Transform or "Feynman's Trick" to show that

$$int_0^infty fraccos(x)-e^-xx,dx=int_0^infty left(fracxx^2+1-frac1x+1right)=0tag3$$

Starting with $(3)$, is tantamount to starting with the real part of $(1)$ and we are done.

So, what are other ways to prove the coveted relationship using real analysis tools only?

Note that for each $x>0$, the improper integral $int_x^inftyfraccos(t)tdt$
converges conditionally (in the sense that $lim_Arightarrowinftyint_x^Afraccos(t)tdt$
exists but $lim_Arightarrowinftyint_x^Aleft|fraccos(t)tright|dt=infty$).
Define $F:(0,infty)rightarrowmathbbR$ by $F(x)=int_x^inftyfraccos(t)tdt$.
Note that we still have $F'(x)=-fraccos(x)x$. For,
begineqnarray*
F'(x) & = & lim_hrightarrow0fracint_x+h^inftyfraccos(t)tdt-int_x^inftyfraccos(t)tdth\
& = & -lim_hrightarrow0fracint_x^x+hfraccos(t)tdth\
& = & -fraccos(x)x.
endeqnarray*

On the other hand, $0$ in the integral $int_0^xfraccos(t)-1tdt$
is a removable singularity. For $tneq0$, we have
begineqnarray*
fraccos(t)-1t & = & frac(1-frac12!t^2+frac14!t^4-cdots)-1t\
& = & -frac12!t+frac14!t^3-frac16!t^5+cdots.
endeqnarray*
Define $phi:mathbbRrightarrowmathbbR$ by $phi(t)=-frac12!t+frac14!t^3-frac16!t^5+cdots$.
It can be proved that the power series converges everywhere (using
root-test), and hence $phi$ is an analytic function. Therefore $int_0^xfraccos(t)-1tdt=int_0^xphi(t)dt$.

Now $fracddxint_0^xfraccos(t)-1tdt=fraccos(x)-1x$.

Finally, define $G:(0,infty)rightarrowmathbbR$ by
$$
G(x)=int_x^inftyfraccos(t)tdt-left[ln(x)+int_0^xfraccos(t)-1tdtright].
$$
Then $G$ is differentiable and $G'(x)=-fraccos(x)x-frac1x-fraccos(x)-1x=0$.
This shows that $G$ is a constant function.