A convergent improper integral of $\ln x/(1+x)^2$

Integration by parts on $[X,1]$. For $0<X<1$, set $u(x)=\ln x$ and $v'(x)=\frac{1}{(1+x)^2}$, so $u'(x)=\frac1x$ and $v(x)=-\frac{1}{1+x}$. Then
$$I(X):=\int_X^1\frac{\ln x}{(1+x)^2}\,dx=\left[-\frac{\ln x}{1+x}\right]_X^1+\int_X^1\frac{dx}{x(1+x)}.$$

First term.
$$\left[-\frac{\ln x}{1+x}\right]_X^1=-\frac{\ln 1}{2}-\left(-\frac{\ln X}{1+X}\right)=\frac{\ln X}{1+X}.$$

Second term. Partial fractions: $\frac{1}{x(1+x)}=\frac1x-\frac1{1+x}$. So
$$\int_X^1\frac{dx}{x(1+x)}=\bigl[\ln x-\ln(1+x)\bigr]_X^1=-\ln 2-\bigl(\ln X-\ln(1+X)\bigr).$$

Combine.
$$I(X)=\frac{\ln X}{1+X}-\ln X+\ln(1+X)-\ln 2=-\frac{X\ln X}{1+X}+\ln(1+X)-\ln 2.$$
(The simplification uses $\frac{1}{1+X}-1=-\frac{X}{1+X}$.)

Limit as $X\to0^+$. Since $X\ln X\to0$ and $\ln(1+X)\to0$, we get
$$I=\lim_{X\to0^+}I(X)=0+0-\ln 2=-\ln 2.$$

Remark. The naive computation
$$\int_0^1\frac{\ln x}{(1+x)^2}\,dx\overset{?}{=}\left[-\frac{\ln x}{1+x}\right]_0^1+\int_0^1\frac{dx}{x(1+x)}$$
is meaningless because $\ln x$ is not defined at $0$ and $\int_0^1\frac{dx}{x}$ diverges. The correct procedure is to integrate over $[X,1]$ with $X>0$ and let $X\to0^+$.