Question:

$$I_n = \left[ (1-x^4)^n \cdot x \right]_0^1 - \int_{0}^{1} n(1-x^4)^{n-1}(-4x^3) \cdot x dx$$ $$= \left[ x(1-x^4)^n \right]_0^1 + 4n \int_{0}^{1} x^4(1-x^4)^{n-1} dx = 0 + 4n \int_{0}^{1} (1-x^4)^{n-1}(1-(1-x^4)) dx$$ $$= 4n \left[ \int_{0}^{1} (1-x^4)^{n-1} dx - \int_{0}^{1} (1-x^4)^n dx \right]$$ $$I_n = 4n I_{n-1} - 4n I_n$$ $$(4n+1) I_n = 4n I_{n-1}$$ $$\frac{I_n}{I_{n-1}} = \frac{4n}{4n+1}$$