= Solution
Write $N_{i,d}$ for the partition-of-unity-normalized degree-$d$ <B-spline>, supported on $[t_i,t_{i+d+1}]$. Assume $d\ge1$. First take distinct knots; endpoint repetitions will be interpreted as confluent limits. For fixed parameter $t$, let $f_t(\tau)=(\tau-t)_+^d$, and form
$$
N_{i,d}(t)=(t_{i+d+1}-t_i)[t_i,\ldots,t_{i+d+1}]f_t,
$$
where the brackets denote a <divided difference> in the knot variable $\tau$. This representation follows from the defining properties, rather than assuming a derivative recurrence. If $t<t_i$, $f_t$ is a degree-$d$ polynomial in all its knot arguments and its order-$(d+1)$ <divided difference> is zero. If $t>t_{i+d+1}$, every truncated-power term is zero. Between knots it is a degree-$d$ polynomial with $d-1$ continuous <derivatives>. Those support and smoothness conditions determine a one-dimensional spline space: there are $d+1$ spans, the continuity conditions leave $2d+1$ coefficients, and vanishing through order $d-1$ at both support ends imposes $2d$ conditions.
The remaining scalar is fixed by the prescribed integral. Put $a=t_i$, $b=t_{i+d+1}$. Integrating the truncated power first gives
$$
\int_a^b(\tau-t)_+^d\,dt=\frac{(\tau-a)^{d+1}}{d+1}\quad(a\le\tau\le b).
$$
The order-$(d+1)$ <divided difference> of this polynomial is its leading coefficient $1/(d+1)$, so the proposed $N_{i,d}$ has integral $(b-a)/(d+1)$, exactly the specified normalization.
Now differentiate with respect to $t$ and apply the final recursion for a <divided difference>:
$$
\begin{aligned}
N_{i,d}'(t)
&=-d(t_{i+d+1}-t_i)[t_i,\ldots,t_{i+d+1}](\tau-t)_+^{d-1}\\
&=d[t_i,\ldots,t_{i+d}](\tau-t)_+^{d-1}
-d[t_{i+1},\ldots,t_{i+d+1}](\tau-t)_+^{d-1}.
\end{aligned}
$$
In terms of normalized lower-degree <B-splines>, this is the <B-spline differentiation formula>
$$
\boxed{N_{i,d}'=\frac d{t_{i+d}-t_i}N_{i,d-1}
-\frac d{t_{i+d+1}-t_{i+1}}N_{i+1,d-1}.}
$$
A zero denominator corresponds to a collapsed-support basis term and contributes zero. Repeated knots follow by coalescing distinct knots; at knots where the classical derivative is discontinuous, read the formula on each open span or one-sided. The stated maximal interior continuity corresponds to simple interior knots, while the clamped endpoint repetitions do not affect the interior argument.
Let the control-point abscissae be the <Greville abscissae>
$$
\xi_i=\frac1d\sum_{r=1}^dt_{i+r},\qquad X(t)=\sum_{i=0}^n\xi_iN_{i,d}(t).
$$
Collect the coefficient of $N_{i,d-1}$ after differentiating:
$$
X'(t)=\sum_{i=1}^n\frac{d(\xi_i-\xi_{i-1})}{t_{i+d}-t_i}N_{i,d-1}(t).
$$
The difference of the two consecutive knot averages telescopes to
$$
\xi_i-\xi_{i-1}=\frac{t_{i+d}-t_i}{d},
$$
so every noncollapsed coefficient is one. The lower-degree basis on the endpoint-trimmed knot vector has <partition of unity>; equivalently the outer lower-degree basis functions on the original vector have collapsed support. Therefore $X'(t)=1$ on the parameter interval. With $d+1$ repeated left endpoint knots at $a$, the clamped basis has $N_{0,d}(a)=1$, all other values zero, and $\xi_0=a$. Thus $X(a)=a$ fixes the constant of integration. Continuity and endpoint limits now give
$$
\boxed{X(t)=t\quad\text{throughout the basic parameter interval}.}
$$
This is <linear precision at Greville abscissae>. It requires neither equal knot spacing nor a restriction on the other control-point coordinates; those determine the remaining coordinates of the <parametric curve>.
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