Differential operator and Integral, are they inverse algebracally? Is this expression a valid one
$$\int dt=1/D$$
Where D is a differential operator, or we can say
$$D=(d/dt)$$
Because we solve questions related to transient analysis so this relationship works well as the voltage-current relationship in capacitor is 
$$v=(1/C) \int(I)dt$$
So we write
$$v=(1/CD)I$$
 A: The short answer is no. The long answer is still no, but longer.
Consider the function $f$ defined by $f(t) = t$. It's derivative is 
\begin{equation}
\dfrac{\mathrm{d}f}{\mathrm{d}t} = 1\;.
\end{equation}
A primitive or indefinite integral of 1 is any function $g$ such that
\begin{equation}
\dfrac{\mathrm{d}g}{\mathrm{d}t} = 1\;.
\end{equation}
This includes $f(t) = t$ but it also includes the function $t\mapsto t+2$, or in fact any function of the form $t\mapsto t+C$, where $C$ is any real number.
So, if you start with at function $f$, compute its derivative then try to go back to $f$, the indefinite integral doesn't know which of the possible functions it must choose. Does it choose $f(t)$ or $f(t) + 2$, etc.? If the integral is the inverse of the derivative, it would have to pick out one of the infinitely many valid primitives of a function. This is actually a very important property: when solving differential equations, the choice of the constant $C$ is specified by the initial conditions. If the indefinite integral is the inverse of the derivative, it would not be possible to adapt the integral to the initial conditions.
If you are familiar with linear algebra there is a nice matrix demonstration. Consider the set of real polynomials of degree 2 or less. Each polynomial can be identified with an element in $\boldsymbol{R}^3$:
\begin{align}
a_0 + a_1x+a_2x^2&&\leftrightarrow&&\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}\;.
\end{align}
For example,
\begin{align}
2 -3x+5x^2&&\leftrightarrow&&\begin{pmatrix}2\\-3\\5\end{pmatrix}\;.
\end{align}
The polynomials $1$, $x$, and $x^2$ form a basis of the vector space of the polynomials of degree 2 or less. The derivative operator can be represented by a $3\times 3$ matrix:
\begin{align}
\dfrac{\mathrm{d}\ }{\mathrm{d}x}(a_0 + a_1x+a_2x^2) = a_1+2a_2x&&\leftrightarrow&&
\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix} =\begin{pmatrix}a_1\\2a_2\\0\end{pmatrix}\;.
\end{align}
The matrix 
\begin{align}
D=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}
\end{align}
is not invertible.
