Integrate

$\begin{aligned} I & = \int_0^\infty \frac{x^4 \ln (x+1)}{(x+1)^7} dx \quad \quad \small \color{#3D99F6}{\text{Let }u = \frac{1}{x+1} \implies x = \frac{1}{u} - 1 = \frac{1-u}{u} \implies dx = - \frac{1}{u^2} du} \\ & = \int_1^0 \left(\frac{1-u}{u} \right)^4 \frac{(-\ln u)u^7}{-u^2} du \\ & = \int_1^0 (u-1)^4 u \ln u \ du \quad \quad \small \color{#3D99F6}{\text{By integration by parts: }f' = (u-1)^4u, \ g = \ln u} \\ & = \left( \frac{u^6}{6} -\frac{4u^5}{5} + \frac{3u^4}{2} - \frac{4u^3}{3} + \frac{u^2}{2} \right) \ln u \ \bigg|_1^0 - \int_1^0 \left( \frac{u^5}{6} -\frac{4u^4}{5} + \frac{3u^3}{2} - \frac{4u^2}{3} + \frac{u}{2} \right) du \\ & = 0 + \frac{u^6}{36} -\frac{4u^5}{25} + \frac{3u^4}{8} - \frac{4u^3}{9} + \frac{u^2}{4} \ \bigg|_0^1 \\ & = \frac{1}{36} -\frac{4}{25} + \frac{3}{8} - \frac{4}{9} + \frac{1}{4} \\ & = \frac{29}{600} \end{aligned}$

$\implies p + q = \boxed{629}$

Using the beta function $B(a,b) \; = \; \int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\,dx$ we see that $\int_0^\infty \frac{x^4 \ln(1+x)}{(1+x)^7}\,dx \; =\; -\frac{\partial B}{\partial b}(5,2) \; = \; \left.\frac{\Gamma(a)}{\Gamma(a+b)}\big[\psi(a+b) - \psi(b)\big] \right|_{(a,b) = (5,2)}$ which is equal to $\frac{\Gamma(5)}{\Gamma(7)}\big[\psi(7) - \psi(2)\big] \; = \; \tfrac{1}{30}\big[\tfrac12 + \tfrac13 + \tfrac14 + \tfrac15 + \tfrac16\big] \; = \; \tfrac{29}{600}$ making the answer $\boxed{629}$ .