Integral 665+1

Integration by parts gives $\begin{aligned} \int_0^X \frac{\ln(1+x) - 1 + e^{-x}}{x^2}\,dx & = \; \left[ -\frac{\ln(1+x) - 1 + e^{-x}}{x}\right]_0^X + \int_0^X \left(\frac{1-e^{-x}}{x} - \frac{1}{1+x}\right)\,dx \\ & = \; -\frac{\ln(1+X) - 1 + e^{-X}}{X} -\ln(1+X) + \int_0^X \frac{1 - e^{-x}}{x}\,dx \\ & = \; -\frac{\ln(1+X) - 1 + e^{-X}}{X} - \ln(1+X) + \Big[(1 - e^{-x})\ln x\Big]_0^X - \int_0^X e^{-x}\ln x\,dx \\ & = \; -\frac{\ln(1+X) - 1 + e^{-X}}{X} - \ln(1+X) + (1 - e^{-X})\ln X - \int_0^X e^{-x}\ln x\,dx \end{aligned}$ for all $X > 0$ , and hence $\int_0^\infty \frac{\ln(1+x) - 1 + e^{-x}}{x^2}\,dx \; = \; -\int_0^\infty e^{-x}\ln x\,dx$ and this second integral is a standard one, so we know that $\int_0^\infty \frac{\ln(1+x) - 1 + e^{-x}}{x^2}\,dx \; = \; \gamma \; = \; \boxed{0.577216}...$