It becomes child's play, part 1 (My seventh integral problem )

$x+1=\sqrt{(x+1)^2}$ $=\sqrt{x^2+x+1+\color{#D61F06}{x}}$ $=\sqrt{x^2+x+1+\color{#D61F06}{\sqrt{x^2}}}$ $=\sqrt{x^2+x+1+\sqrt{x^2+x+1-\color{#624F41}{(x+1)}}}$ $=\sqrt{x^2+x+1+\sqrt{x^2+x+1-\color{#624F41}{\sqrt{(x+1)^2}}}}$ And now the same pattern follows as that from the first step to generate the required pattern. Hence, $\Large \int_{2}^3 (x+1)dx=\dfrac{1}{2}(x^2+2x)|_{\small{2}}^{\small{3}}$ $=\dfrac{7}{2}$ $\huge \therefore 10\times\dfrac{7}{2}=\boxed{\color{#007fff}{35}}$

Let us first try to generalize a few things,

$\sqrt{ a+\sqrt{ a- \sqrt{a+ \sqrt{a-\ldots } }}}=m \\ \sqrt{ a-\sqrt{ a+ \sqrt{a- \sqrt{a+\ldots } }}}=n \\\Rightarrow \begin{cases} \sqrt{a+n}=m \\ \sqrt{a-m}=n \end{cases}$

Squaring both the equations,

$a+n=m^2 \\ a-m=n^2 \\ m^2-n^2=m+n \\ m-n=1$

Now placing $m=\sqrt{a+n}$ in and squaring,

$a+n=n^2+2n+1 \\ n^2+n+1-a=0 \\n=\frac{-1+\sqrt{1-4(1-a)}}{2}$

Similarly,

$m= \frac{1+\sqrt{1-4(1-a)}}{2}$

So, in our original question, $a=x^2+x+1$ and therefore,

$\begin{aligned} n & = \frac{-1+\sqrt{1-4(1-x^2-x-1)}}{2} \\ &= \frac{-1+\sqrt{4x^2+4x+1}}{2} \\ &= \frac{2x}{2}=x \end{aligned}$

Similarly,

$m=x+1$

$\begin{aligned} \displaystyle \int^{3}_{2}(x+1)dx & = \left(\frac{x^2}{2}+x\right) |_{\small{2}}^{\small{3}} \\ & = \frac{3^2}{2}+3-\frac{2^2}{2}-2=\boxed{\frac{7}{2}}\end{aligned}$

Let $\begin{cases} u = \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \cdots }}}} \\ t = \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \sqrt{x^2+x+1 - \sqrt{x^2+x+1 + \cdots}}}} \end{cases}$

$\implies \begin{cases} u^2 = x^2+x+1+t & ...(1) \\ t^2 = x^2+x+1 - u & ...(2) \end{cases}$ $\implies (1)-(2): u^2-t^2 = t+u \implies (u-t)(u+t) = u+t \implies u-t = 1 \implies u = t+1$

$\implies (1): (t+1)^2 = x^2+x+1+ t \implies t^2+2t+1 = x^2 + x + 1 + t \implies t^2+t+1 = x^2 + x + 1 \implies t = x \implies u = x + 1$

Therefore, $\displaystyle \int_2^3 u \ dx = \int_2^3 (x+1) \ dx = \left[\frac {x^2}2 + x \right]_2^3 = \frac {3^2-2^2}2 + 3-2 = 3.5$ $\implies 10V = 10 \times 3.5 = \boxed{35}$ .