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Mathematics Question on Definite Integral

If
n(2n+1)∫01(1−xn)2ndx=1177∫01(1−xn)2n+1dxn(2n+1)\int_{0}^{1}(1−xn)^{2n}dx=1177\int_{0}^{1}(1−x^n)^{2n+1}dx
,then n ∈ N is equal to ______.

Answer

The correct answer is 24
∫01(1−xn)2n+1 dx=∫01(1−xn)2n+1 dx\int_{0}^{1} (1 - x^n)^{2n+1} \,dx = \int_{0}^{1} (1 - x^n)^{2n+1} \,dx
=[(1−xn)2n+1⋅x]01−∫01x⋅(2n+1)(1−xn)2n⋅(−nxn−1) dx= \left[ (1 - x^n)^{2n+1} \cdot x \right]_{0}^{1} - \int_{0}^{1} x \cdot (2n+1)(1 - x^n)^{2n} \cdot (-nx^{n-1}) \,dx
=n(2n+1)∫01(1−(1−xn))(1−xn)2n dx= n(2n+1) \int_{0}^{1} (1 - (1 - x^n))(1 - x^n)^{2n} \,dx
=n(2n+1)∫01(1−xn)2n dx−n(2n+1)∫01(1−xn)2n+1 dx= n(2n+1) \int_{0}^{1} (1 - x^n)^{2n} \,dx - n(2n+1) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx
(1+n(2n+1))∫01(1−xn)2n+1 dx=n(2n+1)∫01(1−xn)2n dx(1 + n(2n+1)) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx = n(2n+1) \int_{0}^{1} (1 - x^n)^{2n} \,dx
(2n2+n+1)∫01(1−xn)2n+1 dx=1177∫01(1−xn)2n+1 dx(2n^2 + n + 1) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx = 1177\int_{0}^{1} (1 - x^n)^{2n+1} \,dx
∴ 2 n 2 + n + 1 = 1177
2 n 2 + n – 1176 = 0
∴n=24or−492\therefore n=24 or −\frac{49}{2}
∴ n = 24