Let $f(x),g(x),h(x)$ are three differentiable functions satisfying: 
Let $f(x),g(x),h(x)$ are three differentiable functions satisfying:
   $\int (f(x)+g(x))dx=\frac{x^3}3+C_1, \int(f(x)-g(x))dx=x^2-\frac{x^3}{3}+C_2$, $\int\frac{f(x)}{h(x)}dx=-\frac1x+C_3$.
  
  $(1)$ The value of $\int (f(x)+g(x)+h(x))dx$ is equal to-
  $(2)$ Number of points of non differentiability of function $\phi(x)=\min\{f(x),f(x)+g(x),h(x)\}$ is equal to-
  
  $(3)$ If number of distinct terms in the expansion of $((1+f(x))+(\frac{f(x)+g(x)}{h(x)}))^{\sum n}$, ($n\in N$) is $31$, then the value of $n$ is-

We get by differentiating first and second equations, $f(x)+g(x)=x^2\tag1$$f(x)-g(x)=2x-x^2\tag2$
So $f(x)=x$ and $g(x)=x^2-x$. And by a similar argument, $h(x)=x^3$. So answer to the first question is $\frac{x^3}{3}+\frac{x^4}{4}+C$. 
For the second one $\phi(x)=\min\{x,x^2,x^3\}$ and all of these are differentiable at every point, and thus, answer to the second question is $0$. 
For the third one, the expression becomes $(1+x+\frac1x)^{\sum n}$, but here is where I got stuck, so how to solve this. And are my previous answers correct?

For the second part, $x=-1,0,1$ are not differentiable as the limit from left and right are not equal.
For the third part, you have $x^0,x^{\pm1},x^{\pm2},...,x^{\pm15}$ so $\Sigma  n=15\implies n=5$.
