Show that $\sum_{r=0}^n\binom{n}{r}\frac{(-1)^r}{2r+1}=\frac{2^{2n}(n!)^2}{(2n+1)!}$

Question

This equation comes from a problem in a textbook, where I was asked to show that $$\int_{0}^{1}(1-x^2)^n\mathrm{d}x=\frac{2^{2n}(n!)^2}{(2n+1)!}$$ I adopt a straightforward approach: $$(1-x^2)^n=\sum_{r=0}^{n}\binom{n}{r}1^{n-r}(-x^2)^r=\sum_{r=0}^{n}\binom{n}{r}(-1)^rx^{2r}$$ and then we can easily get its definite integral, which is $$\sum_{r=0}^{n}\binom{n}{r}\frac{(-1)^r}{2r+1}x^{2r+1}$$ then $$\int_{0}^{1}(1-x^2)^n\mathrm{d}x=\sum_{r=0}^{n}\binom{n}{r}\frac{(-1)^r}{2r+1}x^{2r+1}\bigg]_0^1=\sum_{r=0}^{n}\binom{n}{r}\frac{(-1)^r}{2r+1}$$ and then the remaining problem is to show that $$\sum_{r=0}^n\binom{n}{r}\frac{(-1)^r}{2r+1}=\frac{2^{2n}(n!)^2}{(2n+1)!}$$ I've checked their values by geogebra and it appears that this equation is always true, but how can we preceed?