What is the Cameron-Martin norm associated to $X(t)=\int_0^t B(s) ds+B(t)$? The process $X(t)=\int_0^t B(s) ds+B(t)$ is a centered continuous Gaussian process. Therefore it defines a Gaussian measure on $C[0,T]$. Therefore there is a Cameron-Martin space with Cameron-Martin norm. I can compute the covariance and get some complicated expression.
Is there a clean expression for the Cameron-Martin norm associated to $X$?
 A: I will show that for $x\in\mathcal C^1$, $x$ is in the Cameron-Martin space $\mathcal H$ and
$$ |x|_\mathcal{H}^2 = \int_0^T\left(x'(t)-\int_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds\right)^2\mathrm dt. $$
Expanding the product and using Fubini's theorem, it can be rewritten as
$$ |x|_\mathcal{H}^2 = \int_0^T|x'(t)|^2\mathrm dt + \int_0^T\int_0^Tk(s,t)x'(s)x'(t)\mathrm ds\mathrm dt $$
for some explicit kernel $k$. I claim that with slightly more effort, one can show that $\mathcal H=\mathrm H^1$ as vector spaces, and that the above expression stays true for $x'\in\mathrm L^2$ the weak derivative. Hopefully this is clean enough for your purposes.
The Cameron-Martin space.
Define $\overline F:\mathcal C\to\mathcal C$ as the continuous map sending $B$ to $X$; formally, $F(b):t\mapsto\int_0^tb_sds + b_t$. Set $i_B:\mathrm H^1\to\mathcal C$ the usual Cameron-Martin inclusion for Brownian motion.¹ The important observation is that the continuous map $\overline F\circ i_B:\mathrm H^1\to\mathcal C$ is injective. Let us postpone the proof to the end of the answer.
Define $\mathcal H$ as the image of $\mathrm H^1$ under $\overline F\circ i_B$, together with the inner product coming from $\mathrm H^1$ (we use the injectivity here). In other words, $\mathcal H$ is the set of all $\overline F\circ i_B(b)$ for $b\in\mathrm H^1$, and $|\overline F\circ i_B(b)|_\mathcal{H}^2=|b|_{\mathrm H^1}^2$.  It is obviously canonically isomorphic to $\mathrm H^1$, through some $F:\mathrm H^1\to\mathcal H$.
I claim that the inclusion $i_X:\mathcal H\hookrightarrow\mathcal C$ is the Cameron-Martin space of $X$. Indeed, for any $\phi,\psi\in\mathcal C^*$, and noting that $\overline F\circ i_B=i_X\circ F$,
$$ \begin{align*}
   \mathbb E[\phi(X)\psi(X)]
&= \mathbb E[\phi\circ\overline F(B)\psi\circ\overline F(B)] \\
&= \big\langle\phi\circ\overline F\circ i_B,\psi\circ\overline F\circ i_B\big\rangle_{(\mathrm H^1)^*} \\
&= \big\langle\phi\circ i_X\circ F,\psi\circ i_X\circ F\big\rangle_{(\mathrm H^1)^*} \\
&= \big\langle\phi\circ i_X,\psi\circ i_X\big\rangle_\mathcal{H^*}.
   \end{align*} $$
Expression for the norm.
Suppose $x\in\mathcal C$ is actually of class $\mathcal C^1$, and define
$$ b:t\mapsto
   \int_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds
 = \mathbf e^{-t}\int_0^t\mathbf e^sx'(s)\mathrm ds. $$
Then $b$ is clearly in $\mathrm H^1$, and in fact we will show $x=F(b)$, so $x\in\mathcal H$ and $|x|_\mathcal{H}=|b|_{\mathrm H^1}$ and we have the expected expression for the Cameron-Martin norm.
The fact that $x=F(b)$ follows from computing derivatives:
$$ \frac{\mathrm d}{\mathrm dt}\big(F(b)(t)-x(t)\big)
 = b(t) + b'(t) - x'(t)
 = 0. $$
Injectivity of the transform.
Since $i_B$ is injective, we need only show that $\overline F$ is injective. Choose $b\in\mathcal C$ such that $\overline F(b)=0$. This means precisely that
$$ 0=\overline F(b)(t)=\int_0^tb(s)\mathrm ds + b(t) $$
for all $t$. Then the function
$$ t\mapsto\mathbf e^t\int_0^tb(s)\mathrm dt $$
is $\mathcal C^1$ with vanishing derivative hence always zero, and using the relation again,
$$ b(t) = -\int_0^tb(s)\mathrm ds = -\mathbf e^{-t}\cdot0 = 0. $$

¹ $\mathrm H^1$ is the Sobolev space of functions $b$ that can be written as the integral of a function $\dot b\in\mathrm L^2$.
