Does this equality always hold? Is it true in general that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x} f(u,x) \mathrm{d}u = \int_0^{x} \left( \frac{\mathrm{d}}{\mathrm{d}x} f(u,x) \right)\mathrm{d}u +f(x,x )$ ?
Thank you for your help!

Yes, it is, under the conditions indicated below.
Let $$I(x)=\displaystyle\int_{0}^{x}f(u,x)\; \mathrm{d}u.\qquad(\ast)$$
If $f(u,x)$ is a continuous function and  $\partial f/\partial x$ exists and is continuous, then  
$$I^{\prime }(x)=\displaystyle\int_{0}^{x}\dfrac{\partial f(u,x)}{\partial x}\; \mathrm{d}u+f(x,x)\qquad(\ast\ast)$$
follows from  the Leibniz rule and chain rule. 
Note: the integrand of $(\ast\ast)$ is a partial derivative.
It generalizes to the integral 
$$I(x)=\displaystyle\int_{u(x)}^{v(x)}f(t,x)\; \mathrm{d}t.$$
Under suitable conditions ($u(x),v(x)$ are differentiable functions, $f(t,x)$ is a continuous function and  $\partial f/\partial x$ exists and is continuous), we have 
$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}\; \mathrm{d}t+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$
