Who's up to the challenge?15

The integral is equal to $F'(5)$ , where $F(a) \; = \; \int_0^\infty \frac{x^{a-2}}{(x^2+1)^a}\,dx \; =\; \tfrac12\int_0^\infty \frac{u^{\frac12(a-1)}}{(1+u)^a}\,da \; = \; \tfrac12B\big(\tfrac12(a-1),\tfrac12(a+1)\big)$ and so $F'(5) \; =\;F(5)\big[\tfrac12\psi(2) + \tfrac12\psi(3) - \psi(5)\big] \; = \; \tfrac{1}{24}\big[\tfrac12H_1 + \tfrac12H_2 - H_4\big] \; = \; \tfrac{1}{24}\big[H_2 - H_4 - \tfrac14\big]$ making the answer $2 + 4 + 1 + 4 + 24 \,=\, \boxed{35}$ , but why give the answer like this? The quantity $F'(5)$ is equal to $-\tfrac{5}{144}$ . You could even keep the answer the same by saying that the integral is of the form $-\tfrac{A}{B}$ where $A,B$ are coprime integers, and ask for $\tfrac{B+31}{A}$ (or something similar) as the answer.