Integration Grandmaster - P2

The substitution $y \; =\; \frac{x\sinh x + \cosh x}{x\cosh x - \sinh x}$ gives $\frac{dy}{dx} \; = \; \frac{x\cosh x + 2\sinh x}{x\cosh x - \sinh x} - \frac{x\sinh x(x\sinh x + \cosh x)}{(x \cosh x - \sinh x)^2} \; =\; \frac{x^2 - 2\sinh^2x}{(x\cosh x - \sinh x)^2}$ and hence $\frac{1}{y}\,\frac{dy}{dx} \; =\; \frac{x^2 - 2\sinh^2x}{(x\cosh x - \sinh x)(x\sinh x + \cosh x)} \; = \; \frac{x^2 - 2\sinh^2x}{(x^2-1)\sinh x \cosh x + x}$ so that $F(x) \; = \; \int \frac{1}{2y}\,dy \; = \; \tfrac12\ln y + c$ With the given value of $F(\ln 2)$ , we deduce that $c=0$ , and hence $F(x) \; =\; \tfrac12\ln\left(\frac{x\sinh x + \cosh x}{x\cosh x - \sinh x}\right)$ which implies that $F(2) \; =\; \tfrac12\ln\left(\frac{3e^4 - 1}{e^4 + 3}\right)$ and hence $\lfloor 10000 F(2)\rfloor = \boxed{5194}$ .