Find f(x) $\int_0^x f(u)du - f'(x) = x$ Find f(x)
$$\int_0^x f(u)du - f'(x) = x$$
I was not given f(0) which makes it difficult for me to find f(x). This is what I have thus far: 
$$\frac{F(p)}{p}-pF(p)+f(0)=\frac{1}{p^2}$$
$$\frac{F(p)}{p}-pF(p)=\frac{1}{p^2}-f(0)$$
$$F(p)(\frac{1}{p}-p)=\frac{1}{p^2}-f(0)$$
$$F(p)=\frac{\frac{1}{p^2}-f(0)}{(\frac{1}{p}-p)}$$
Can I even take the Inverse Laplace to get f(x) or do I have to do something entirely different. Once again I was not given f(0). And I used the formula:
$$L(\int_0^t f(u)du)=\frac{1}{p} F(p)$$
where L=Laplace Transform
 A: A start: Differentiate. By the Fundamental Theorem of Calculus,we have $f(x)-f''(x)=1$. Solve this differential equation.
By substituting in the original equation, we can see  that $f'(0)=0$.
A: Edit: I add done a mistake in my previous answer (taking $f(0)=0$ instead of $f'(0)=0$). It is corrected in this version:
Taking $x=0$, one finds 
$$f'(0)=0  \ \ (1)$$ 
Thus this integro-differential equation is equivalent, by derivation, to
$$f(x)-f''(x)=1 \ \ \text{with condition (1)}$$ 
which has solution 
$$1+Ae^{x}+Be^{-x} \ \ \text{with condition (1)}$$
(particular solution of the diff. equ. + general solution of the homogeneous associated diff. equ. $f(x)-f''(x)=0$.)
Taking (1) into account, and because $f'(x)=Ae^{x}-Be^{-x}$, we have $f'(0)=A-B=0$; thus, the solution to the question is:
$$f(x)=1+Ae^{x}+Ae^{-x}=1+a\cosh{x}$$
$a$ being an arbitrary real constant. 
Of course, treating the initial equation by Laplace Transform yields the same results (as confirmed by the result of @Jan Eerland)
A: $$\int_{0}^{x}f(u)\space\text{d}u-f'(x)=x\Longleftrightarrow$$
$$\mathcal{L}_{x}\left[\int_{0}^{x}f(u)\space\text{d}u-f'(x)\right]_{(s)}=\mathcal{L}_{x}\left[x\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_{x}\left[\int_{0}^{x}f(u)\space\text{d}u\right]_{(s)}-\mathcal{L}_{x}\left[f'(x)\right]_{(s)}=\mathcal{L}_{x}\left[x\right]_{(s)}\Longleftrightarrow$$
$$\frac{f(s)}{s}-sf(s)+f(0)=\frac{1}{s^2}\Longleftrightarrow f(s)\left[\frac{1}{s}-s\right]=\frac{1}{s^2}-f(0)\Longleftrightarrow$$
$$f(s)=\frac{\frac{1}{s^2}-f(0)}{\frac{1}{s}-s}\Longleftrightarrow f(s)=\frac{1-f(0)s^2}{s-s^3}\Longleftrightarrow$$
$$\mathcal{L}_{s}^{-1}\left[f(s)\right]_{(x)}=\mathcal{L}_{s}^{-1}\left[\frac{1-f(0)s^2}{s-s^3}\right]_{(x)}\Longleftrightarrow f(x)=1+(f(0)-1)\text{cosh}(x)$$
