Integrate terms in AP

Proposition

$\int {x^n (x^n + d) {(x^n + 2d)}^{1/n} \ dx} = \frac{1}{2(n+1)} x^{n+1}{(x^n + 2d)}^{1/n}$

Proof

Put $\color{#D61F06} x^n + d = t$ ,

$= \frac{1}{n} \int{ {(t-d)}^{1/n} t {(t+d)}^{1/n} \ dt}$

$= \frac{1}{n} \int{ t {(t^2 - d^2)}^{1/n} \ dt}$

Put $\color{#D61F06} t^2 - d^2 = u^n$ ,

$= \int{\frac{u^n}{2}\ du}$

$= \frac{u^{n+1}}{2(n+1)}$

$= \frac{x^{n+1} {(x^n + 2d)}^{1/n + 1}}{2(n+1)}$

Consider ,

$\begin{aligned} \large \int{x^n(x^n+d)(x^n+2d)^{\tfrac{1}{n}} dx}\end{aligned}$

it can be rewritten as,

$\begin{aligned} \large \int{x^{n-1}(x^n+d)(x^{2n}+2dx^n)^{\tfrac{1}{n}} dx}\end{aligned}$

now let , $\begin{aligned} \large t&=\large x^{2n}+2dx^n\\\large dt&=\large 2n x^{2n-1}+2ndx^{n-1}\\ &=\large 2n\times x^{n-1}(x^n+d) \end{aligned}$

The given integral becomes,

$\begin{aligned} \large \frac{1}{2n}\int{t^{\tfrac{1}{n}}dt}&=\dfrac{t^{\tfrac{1}{n}+1}}{2n( \dfrac{1}{n}+1)}+C\\ &=\large \dfrac{(x^{2n}+2dx^n)^{\tfrac{1}{n}+1}}{2(n+1)}+C\\ &=\dfrac{x^{n+1}\times(x^{n}+2d)^{\tfrac{1}{n}+1}}{2(n+1)}+C\end{aligned}$

$\begin{aligned} \large \int_{0}^p{x^{n-1}(x^n+d)(x^{2n}+2dx^n)^{\tfrac{1}{n}} dx}=\dfrac{p^{n+1}\times(p^{n}+2d)^{\tfrac{1}{n}+1}}{2(n+1)}\end{aligned}$

in the above problem $n=2016 ,d=1008$ and $p=1$ ,on simplifying we get

$\begin{aligned} I&=\dfrac{1\times {2017}^{\tfrac{1}{2016}+1}}{4034}\\&=\dfrac{1\times {2017}^{\tfrac{1}{2016}}}{2}\end{aligned}$

$\Rightarrow a=2017,b=2016,c=2$

$b+c-a=2016+2-2017=\boxed{1}$

Since $\begin{aligned} x^n(x^n+d)(x^n + 2d)^{\frac1n} & = \tfrac12\Big[x^n\big(x^n + (x^n + 2d)\big)(x^n + 2d)^{\frac1n}\Big] \; = \; \tfrac12\Big[x^{2n}(x^n + 2d)^{\frac1n} +x^n(x^n + 2d)^{\frac{n+1}{n}} \Big] \\ &= \tfrac12\Big[x^{n+1} \times x^{n-1}(x^n + 2d)^{\frac1n} + x^n \times (x^n + 2d)^{\frac{n+1}{n}}\Big] \; = \; \frac1{2(n+1)}\frac{d}{dx}\Big[x^{n+1}(x^n + 2d)^{\frac{n+1}{n}}\Big] \end{aligned}$ we have $\int_0^1 x^n(x^n+d)(x^n + 2d)^{\frac1n} \,dx \; =\; \frac{1}{2(n+1)} (2d+1)^{\frac{n+1}{n}}$ Putting $n=2d$ we obtain $\tfrac12(2d+1)^{\frac{1}{2d}}$ , making the answer $2d + 2 - (2d+1) = \boxed{1}$ .