Purpose Of Adding A Constant After Integrating A Function I would like to know the whole purpose of adding a constant termed constant of integration everytime we integrate an indefinite integral $\int f(x)dx$. I am aware that this constant "goes away" when evaluating definite integral $\int_{a}^{b}f(x)dx $. What has that constant have to do with anything? Why is it termed as the constant of integration? Where does it come from?
The motivation for asking this question actually comes from solving a differential equation $$x \frac{dy}{dx} = 5x^3 + 4$$ By separation of $dy$ and $dx$ and integrating both sides, $$\int dy = \int\left(5x^2 + \frac{4}{x}\right)dx$$  yields $$y = \frac{5x^3}{3} + 4 \ln(x) + C .$$
I've understood that $\int dy$ represents adding infinitesimal quantity of $dy$'s yielding $y$ but I'am doubtful about the arbitrary constant $C$.

There are many great answers here, but I just wanted to chime in with my favorite example of how things can go awry if one forgets about the constant of integration.
Consider 
$$\int \sin(2x) dx.$$
We will find antiderivatives in two ways. First, a substitution $u=2x$ yields:
$$\int \frac{\sin(u)}{2}du = -\frac{\cos(u)}{2} = -\frac{\cos(2x)}{2}.$$
Second, we use the identity $\sin(2x)=2\sin(x)\cos(x)$ and a substitution $v=\sin(x)$:
$$\int \sin(2x)dx = \int 2\sin(x)\cos(x)dx = \int 2vdv =v^2 = \sin^2(x).$$
Thus, we have found two antiderivatives of $\sin(2x)$ that are completely different! Namely
$$F(x)=\sin^2(x) \quad \text{and} \quad G(x)=-\frac{\cos(2x)}{2}.$$
Notice that $F(x)=\sin^2(x)\neq -\cos(2x)/2=G(x)$. For instance, $F(0)=\sin^2(0) = 0$ but $G(0)=-\cos(2\cdot 0)/2=-1/2$. So, what happened?
We forgot about the constant of integration, that's what happened. The theory of integration tells us that all antiderivatives differ by a constant. So, if $F(x)$ is an antiderivative, then any other antiderivative $G(x)$ can be expressed as $G(x)=F(x)+C$ for some constant $C$. In particular, our antiderivatives above must differ by a constant. Indeed, the constant $C$ in this case is exactly $C=-\frac{1}{2}$:
$$F(x)+C = F(x) - \frac{1}{2} = \sin^2(x)-\frac{1}{2} = \frac{(1-\cos(2x))}{2}-\frac{1}{2} = -\frac{\cos(2x)}{2} = G(x),$$
where we have used the trigonometric identity $\sin^2(x) = (1-\cos(2x))/2.$ 
