How to prove $\int_{0}^{1} \left( \ln \left( 1 + \frac{1}{x} \right) - \frac{1}{1+x} \right) \ \mathrm dx$ converges

It's worth pointing out that it's not really necessary to find the anti-derivative of the integrand in order to establish that the improper integral converges. It suffices to note that

$$\ln\left(1+{1\over x}\right)-{1\over1+x}=\ln(x+1)-\ln x-{1\over1+x}$$

and that $\ln(x+1)$ and $1/(1+x)$ are continuous (hence integrable) on $[0,1]$, so the convergence of the integral comes down to the convergence of $\int_0^1\ln x\,dx$. Even this doesn't require knowing the anti-derivative of $\ln x$, provided you have an inequality such as $|\ln x|\lt2/\sqrt x$ for $0\lt x\lt1$ (which is easy to show by showing that $f(x)=2x^{-1/2}+\ln x$ is decreasing on $(0,1)$ and noting that $f(1)=2$ is positive), so that

$$\int_0^1|\ln x|\,dx\lt\int_0^1{2\over\sqrt x}\,dx=4\sqrt x\,\big|_0^1=4$$

The value in pointing this out is that not every integrand is easy to integrate in closed form, but in many cases the source of the impropriety in an improper integral can be identified, split off, and dealt with by comparing it to something that is easy to integrate. In short, to prove that an improper integral converges, you don't have to find its exact value, you just have to show it's bounded.

$\lim_{t \to \infty} \frac {\ln (1+t)} t=0$ by L'Hopital's Rule. So $\lim_{x \to 0+} x \ln (1+\frac 1x) =0$. Hence, the answer is $(1)(\ln 2)-0=\ln 2$.