Long road ahead

$\begin{aligned} f(x) & = (x+1)(x+2)(x+3)\cdots(x+2014) \\ f'(x) & = (x+2)(x+3)(x+4)\cdots(x+n) + (x+1)(x+3)(x+4)\cdots(x+n) + \cdots + (x+1)(x+2)(x+3)\cdots(x+2013) \\ & = f(x) \left(\frac 1{x+1} + \frac 1{x+2} + \frac 1{x+3} + \cdots + \frac 1{x+2014} \right) \\ & = f(x) \sum_{k=1}^{2014} \frac 1{x+k} \end{aligned}$

$\displaystyle \implies \frac {f'(x)}{f(x)} = \sum_{k=1}^{2014} \frac 1{x+k}$

Then, we have:

$\begin{aligned} I & = \int_{x+1}^{x+2} \frac {f'(t)}{f(t)} dt = \int_{x+1}^{x+2} \sum_{k=1}^{2014} \frac 1{t+k} dt = \sum_{k=1}^{2014} \ln (t+k) \bigg|_{x+1}^{x+2} \\ & = \sum_{k=1}^{2014} (\ln (x+k+2) - \ln(x+k+1)) \\ & = \ln (x+2016) - \ln (x+2) \end{aligned}$

Now, $I = \ln (x+2016) - x^2 = \ln (x+2016) - \ln (x+2)$ , $\implies x^2 = \ln (x+2)$ . Note that $x^2$ is decreasing when $x<0$ and increasing when $x>0$ and $\ln (x+2)$ is increasing for $x>-2$ . When $x=-1$ , $x^2 > \ln (x+2)$ ; $x=0$ , $x^2 < \ln (x+2)$ ; and $x=2$ , $x^2 > \ln (x+2)$ . Therefore, the equation has two roots, one negative and one positive, $\implies \boxed{n=1, m=1}$ .

First of all, we have $f(x)=(x+1)(x+2)(x+3)...(x+2014)$ . Then, we calculate the integral by substituting $f(t)=\phi$ , obtaining $\displaystyle \int_{f(x+1)}^{f(x+2)} \frac{1}{\phi} d\phi = \ln(f(x+2))-\ln(f(x+1)) = \ln[\frac{(x+3)(x+4)(x+5)...(x+2016)}{(x+2)(x+3)(x+4)...(x+2015)}] = \ln(x+2016)-\ln(x+2)$ . Now that we have calculated the integral, we substitute the result in the given equation, obtaining: $\ln(x+2016)-\ln(x+2)=\ln(x+2016)-x^{2} \rightarrow x^{2}-\ln(x+2)=0$ . We take a new function $g:(-2,\infty)\rightarrow\mathbb{R}$ , $g(x)=x^{2}-\ln(x+2)$ , which is derivable on $(-2,\infty)$ . By solving the equation $g'(x)=0$ , we get the solutions $x_{1}=\frac{-2-\sqrt{6}}{2}$ and $x_{2}=\frac{-2+\sqrt{6}}{2}=0.225$ . From the solutions, only $x_{2}$ belongs to the given interval. Since, $g'(x)=\frac{2x^{2}+4x-1}{x+2}$ and only $x_{2}$ belongs to $(-2,\infty)$ , we notice that $g'(x)<0$ for $x<x_{2}$ and $g'(x)\geq0$ for $x\geq x_{2}$ , therefore $x_{2}$ is an absolute minimum for the function $g$ . Next, we calculate $\lim_{x \to -2} g(x) = +\infty$ and $\lim_{x \to \infty}= +\infty$ . For the next part, we simply calculate $g(x)$ for different values of x: $g(-1)=1$ , $g(0)=-\ln2=-0.7$ , $g(1)=1-\ln3<0$ , $g(2)=4-\ln6>0$ . Therefore, we conclude that $-1<s_{1}<0$ and $1<s_{2}<2$ , where $s_{1}$ and $s_{2}$ and the solutions to the original equation. The correct answer is $n=1$ , $m=1$ .