Using the identity
$$\ln \Gamma(x+1) = \left(x+\frac{1}{2}\right)\ln x - x + \ln\sqrt{2\pi} - \int_{0}^{+\infty} \frac{h(u)}{u+x} du$$
show that for all strictly positive real $x$,
$$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)} = \ln x + \frac{1}{2x} + \int_{0}^{+\infty} \frac{h(u)}{(u+x)^{2}} du$$