How to find the inverse operator? Let $A: C([0,1])\to C([0,1])$ be a linear operator defined with:
$A(x(t)) = x(t) + \int_0^t x(s)\,ds$.
It is actually easy to see that $A$ injective is, but it a bit of problem to show that it's bijective. Though, the main puzzle is: how to find the inverse operator? I myself tried but didn't far away of the expression 
$$
A^{-1} (x(t)) = A^{-1} \int_0^t x(s)\,ds - x(t)
$$
Thanks in forward!
 A: Consider $A$ to be a perturbation of the identity operator:  $A = I - B$, where 
$$
B(x)(t) := -\int\limits_{0}^{t} x(s) \, ds.
$$
If we can prove that the Neumann series
$$
\tag{$*$}
\sum\limits_{k=0}^{\infty} B^k
$$
converges in the operator norm, we are done: then 
$$
A^{-1} = (I - B)^{-1} = \sum\limits_{k=0}^{\infty} B^k.
$$
We have the following equality
$$
\lVert B^k \rVert = \frac{1}{k!}
$$
(for a proof, see, e.g., Norm of integral operator), hence the Neumann series $(*)$ is convergent.
A: If $f=Ax$ and we assume for the moment that $x$ is differentiable, from 
$$
f(t)=x(t)+\int_0^t x(s)\,dx,
$$
we get 
$$
f'=x'+x.
$$
This is a linear differentiable equation on $x$, with solution 
$$
x(t)=x(0)e^{-t}+e^{-t}\int_0^te^sf'(s)\,ds.
$$
Of course this only works if $f$ is differentiable. But, using parts, 
$$
\int_0^te^s f'(s)\,dx=e^tf(t)-f(0)-\int_0^te^sf(s)\,ds, 
$$
so we may write 
$$\tag1
x(t)=x(0)e^{-t}+f(t)-e^{-t}f(0)-e^{-t}\int_0^te^sf(s)\,ds
$$
Now, $Ax(0)=x(0)$, so $x(0)=f(0)$ above. So $(1)$ says that 
$$\tag2
\boxed{(A^{-1}y)(t)=y(t)-e^{-t}\int_0^te^sy(s)\,ds. }
$$
It is now an easy exercise to check that $(2)$ is actually the inverse of $A$, for all $x\in C[0,1]$. 
