Is $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$ equal to $\frac{1}{2n+2}$? On p. 146 of Principles of Mathematical Analysis, Rudin claims that $\int_{0}^{1} x(1-x^2)^ndx=\frac{1}{2n+2}$. But evaluating the integral I found it equal to $$\int_{0}^{1}x(\sum\limits_{k=0}^{n}{{n}\choose{k}}(-x^2)^k)dx=\int_{0}^{1}\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k+1}dx=\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}\frac{x^{2k+2}}{2k+2}$$ evaluated at $x=1$, which equals $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$. So now I'm just wondering, why is this equal to $\frac{1}{2n+2}$?

$$\begin{align*}
\sum_{k=0}^n\binom{n}k\frac{(-1)^k}{2k+2}&=\frac{1}2\sum_{k=0}^n\frac{1}{k+1}\binom{n}k(-1)^k\\
&=\frac{1}2\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}(-1)^k\\
&=\frac{1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^{k-1}\\
&=\frac{-1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}\\
&=\frac{-1}{2n+2}\left(\sum_{k=0}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}-\binom{n+1}0(-1)^01^{n+1}\right)\\
&=\frac{-1}{2n+2}\left((-1+1)^{n+1}-1\right)\\
&=\frac{1}{2n+2}
\end{align*}$$
by the binomial theorem.
