The integer part of a telescoping sum

Telescoping. Since $x_{k+1}=x_k(1+x_k)$,
$$\frac1{x_k}-\frac1{x_{k+1}}=\frac{x_{k+1}-x_k}{x_kx_{k+1}}=\frac1{1+x_k}.$$
Hence
$$\sum_{k=1}^{2025}\frac1{1+x_k}=\frac1{x_1}-\frac1{x_{2026}}=2-\frac1{x_{2026}}.$$

Size of $x_{2026}$. The sequence is positive and increasing ($x_{n+1}-x_n=x_n^2>0$). Moreover $x_2=\frac34$ and $x_3=\frac34\cdot\frac74=\frac{21}{16}>1$. So $x_{2026}>x_3>1$, hence $0<\frac1{x_{2026}}<1$.

Conclusion. The sum lies in $(1,2)$, so its integer part is $\boxed1$.

Check. With only $6$ terms the sum is $2-\frac1{x_7}\approx1.99996$.

Lesson. When a recurrence has the form $x_{n+1}=x_n(x_n+c)$, the quantity $\frac1{x_n+c}$ telescopes.