Proving that $\int_{0}^{1}x^{n-1}(1-x)^n \mathrm dx =\frac{1}{n {2n\choose n}}$ I am working on the following problem and I've arrived at an integral which I need to show equals the proposition.

Prove that $\displaystyle \sum_{k=0}^{k=n}\dfrac{(-1)^{k}{n\choose k}}{n+k}=\dfrac{1}{n{2n\choose n}}$.


My Attempt:
Let us consider for the sake of convenience the notation ${n\choose k}=\text{C}_k$
$$\begin{aligned}(1-x)^n&=\text{C}_0-\text{C}_1x+\text{C}_2x^2-\text{C}_3x^3+\ldots+(-1)^{n}\text{C}_nx^n\\ \int _{0}^{1}x^{n-1}(1-x)^n \mathrm dx&=\int_{0}^{1}\text{C}_0x^{n-1}-\text{C}_1x^n+\text{C}_2x^{n+1}+\ldots+(-1)^{n}\text{C}_nx^{2n-1}\mathrm dx\\ \int_{0}^{1}x^{n-1}(1-x)^n \mathrm dx&=\sum_{k=0}^{k=n}\dfrac{(-1)^{n}\text{C}_k}{n+k}\end{aligned}$$

I now need  to prove that the integral equals $\left[{2n\choose n}\right]^{-1}$. But I am unclear on how to proceed with simplifying the integral. I would appreciate any hints apart from usage of Integration by Parts technique. Thanks
 A: Let's put
$$I_{k,n} = \int_0^1 x^k (1-x)^n dx.$$
By integration by parts we have:
$$I_{k,n} = \frac{n}{k+1} I_{k+1, n-1}. $$
Now we can iterate this to get
$$ I_{k,n} = \frac{n}{k+1} I_{k+1,n-1} = \frac{n}{k+1} \frac{n-1}{k+2} I_{k+2, n-2} = \ldots = \frac{n!}{(k+1)(k+2) \ldots (k+n)} I_{k+n,0}. (*)$$
We have $$I_{k+n,0} = \int_0^1 x^{k+n} dx = \frac{1}{k+n+1}.$$
Now it remains to write the fraction from $(*)$ as a binomial coefficient.
A: $$I=\int_{0}^{1}x^{n-1}(1-x)^n dx= \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} =\frac{(n-1)! n!}{(2n)!}=\frac{1}{n{2n \choose n}}$$
A: $$\dfrac{d(x^m(1-x)^n)}{dx}=mx^{m-1}(1-x)^n-nx^m(1-x)^{n-1}$$
Integrate both sides wrt $x,$ $$x^m(1-x)^n=mI(m-1,n)-nI(m,n-1)$$ where $$I(m,n)=\int x^m(1-x)^n\ dx,J(m,n)=\int_0^1x^m(1-x)^n\ dx$$
and $$J(m-1,n)=\dfrac nm\cdot J(m,n-1)$$
$$=\dfrac{n(n-1)}{m(m-1)}\cdot J(m+1,n-2)$$
$$=\cdots$$
$$=\dfrac{n(n-1)\cdots(n-r+1)}{m(m-1)\cdots(m-r+1)}\cdot J(m+r-1,n-r)$$
$r=n\implies$ $$J(m-1,n)=\dfrac{n!
}{m(m-1)\cdots(m-n+1)}\cdot J(m+n-1,0)$$
$\displaystyle J(m+n-1,0)=\dfrac1{m+n}$
Finally set $m=n$
A: Not sure if this helps,
Let $$I = \int_{0}^{1}x^{n-1}(1-x)^{n}dx = \int_{0}^{1}(1-x)^{n-1}x^{n}dx$$
Therefore $$2I = \int_{0}^{1}x^{n-1}(1-x)^{n-1}dx$$
Let, $$x = sin^2(u) $$
$$dx = 2sin(u)cos(u)du$$
$$2I = \int_{0}^{\pi/2}2sin^{2n-1}(u)cos^{2n-1}(u)du$$ for which standard results exist
