Some other non-trivial sequence in disguise

Since $\int_0^1 x^k \ln x\,dx \; = \; -\frac{1}{(k+1)^2} \hspace{2cm} k \ge 0$ we see that $J_r \; = \; -\sum_{k=0}^{r-1} \int_0^1 x^k \ln x\,dx \; = \; -\int_0^1 \frac{1-x^r}{1-x} \ln x\,dx \hspace{2cm} r \in \mathbb{N}$ and hence, for $N \in \mathbb{N}$ , $\begin{aligned} K_N & = \sum_{r=1}^N \binom{N}{r} (-1)^{r-1}J_r \; = \; \int_0^1 \sum_{r=1}^N \binom{N}{r}(-1)^r (1-x^r) \frac{\ln x}{1-x}\,dx \\ & = \int_0^1 \left\{ (1-1)^N - 1 - (1-x)^N + 1\right\} \frac{\ln x}{1-x}\,dx \; =\; -\int_0^1 (1-x)^{N-1} \ln x \,dx \\ & = -\int_0^1 x^{N-1} \ln(1-x)\,dx \; = \; \int_0^1 x^{N-1} \left(\sum_{j=1}^\infty \frac{1}{j}x^j\right)\,dx \\ & = \sum_{j=1}^\infty \frac{1}{j(j+N)} \; = \; \frac{1}{N}\sum_{j=1}^\infty \left(\frac{1}{j} - \frac{1}{j+N}\right) \\ & = \frac{1}{N}H_N \end{aligned}$ where $H_N$ is the $N$ th Harmonic number. We want $K_{10} = \tfrac{1}{10}H_{10} = \frac{7381}{25200}$ , making the answer $\boxed{32581}$ .