If $$\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x$$, then $\mathrm{n} \in \mathbf{N}$ is equal to ______________.

<p>$$\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} } $$</p> <p>$$ = \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx} $$</p> <p>$= n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx}$</p> <p>$$ = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx - n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $$</p> <p>$$(1 + n(2n + 1))\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx} } $$</p> <p>$$(2{n^2} + n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = 1177\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $$</p> <p>$\therefore$ $2{n^2} + n + 1 = 1177$</p> <p>$2{n^2} + n - 1176 = 0$</p> <p>$\therefore$ $n = 24$ or $- {{49} \over 2}$</p> <p>$\therefore$ $n = 24$</p>