Integration Grandmaster - P3

Using the substitution $y = \ln(x+1)$ , then integral becomes $A \; = \; \int_0^1 \frac{(e^y - 1)^2 \sin^2y}{(e^{\frac12y}+1)^2}e^y\,dy \; = \; \int_0^1 \big(e^{\frac12y}-1\big)^2e^y\,\sin^2y\,dy \; = \; J(2) - 2J(\tfrac32) + J(1)$ where $J(a)$ can be evaluated by standard integration tricks (double angles, integration by parts): $J(a) \; = \; \int_0^1 e^{ay}\sin^2y\,dy \; = \; \frac{1}{a(4+a^2)}\Big[e^a(2 + a^2\sin^21 - a\sin 2) - 1\Big]$ and hence $A = \frac{1}{600}\Big[16e^{\frac32}(9\cos 2 + 12\sin 2 - 25) + 60e(\cos 2 + 2\sin 2 - 5) - 75e^2(\cos 2 + \sin 2 - 2) - 59\Big]$ which makes $\lfloor 10000 A \rfloor = \boxed{1400}$ .