Solution
= Solution
Differentiation under the integral sign gives
$$
U'(x)=-\int_0^\infty\frac{te^{-xt}}{1+t}\,dt,
\qquad
U''(x)=\int_0^\infty\frac{t^2e^{-xt}}{1+t}\,dt.
$$
Hence
$$
\begin{aligned}
xU''+(1-x)U'-U
&=\int_0^\infty e^{-xt}
\frac{xt^2-(1-x)t-1}{1+t}\,dt\\
&=\int_0^\infty e^{-xt}(xt-1)\,dt\\
&=x\frac1{x^2}-\frac1x=0.
\end{aligned}
$$
Thus
$$
\boxed{xU''(x)+(1-x)U'(x)-U(x)=0}.
$$