why do we put arbitrary constant in integration
It's...
The fundamental theorem of calculus:
$\displaystyle F(x)=\int_a^x{f(t)dt}$
where $\displaystyle f$ is a real-valued function continuous on $\displaystyle [a, b]$. Then, $\displaystyle F$ is continuous on $\displaystyle [a, b]$, differentiable on the open interval $\displaystyle (a, b)$, and:
$\displaystyle F'(x)=f(x)$ $\displaystyle \\\\$ $\displaystyle \forall x \in (a,b)$
Say you want to integrate $\displaystyle f(x) = x^2$. Writing $\displaystyle \int{2x}\;{dx} = x^2$ means to say that the function $\displaystyle f(x) = 2x$ satisfies $\displaystyle F'(x) = x^2$. But is it the only one that satisfies this relation? Clearly, $\displaystyle 2x+1$ satisfies; so do $\displaystyle 2x+2$ and $\displaystyle 2x+3$. In fact, for any number $\displaystyle C$, it's true that $\displaystyle 2x+C$ satisfies the relation. Finding $\displaystyle \int{f(x)}\;{dx} $ means finding the set of all function which satisfy F'(x) = f(x), i.e the set of all anti-derivatives of f(x).