How to evaluate: $\int_0^1x^{n-1}(1-x)^{n+1}dx$ How can I evaluate the following integral? ($n \in R$, $n>0$)
$$\int_0^1x^{n-1}(1-x)^{n+1}dx$$

I was solving the following problem (as practice) in school:
Prove that the sum of $n+1$ terms of $$\frac{C_0}{n(n+1)} - \frac{C_1}{(n+1)(n+2)} + \frac{C_2}{(n+2)(n+3)}- \cdot\cdot\cdot = \int_0^1x^{n-1}(1-x)^{n+1}dx$$

Wolfram Alpha says that the integral evaluates to:
$$\frac{\Gamma(n)\Gamma(n+2)}{\Gamma(2n+2)}=\int_0^1x^{n-1}(1-x)^{n+1}dx$$

To reiterate... :
1) How can I evaluate the indefinite integral of the above version?
2) How can I evaluate the definite integral?
3) How to prove the LHS - RHS equality in the aforementioned problem?
 A: There's not a lot that can be said with respect to most of your questions here without appealing to special functions: necessary because of the integral and functions involved, but perhaps unsatisfying as a result.
The most important function involved here is the beta function, defined by
$$\text B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}dt$$
This has a close relationship with the gamma function $\Gamma(x)$ (a generalization of the factorial to any complex number that is not a non-positive integer) as
$$\text B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
First we'll answer your simpler question however.




*How can I evaluate the definite integral?


You have
$$\mathcal I := \int_0^1x^{n-1}(1-x)^{n+1}dx$$
A simple tweak of rewriting $n+1 = (n+2)-1$ allows us to claim that
$$\mathcal I = \text B(n,n+2) = \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(2n+2)}$$
To see this, just let $x=n,y=n+2$ in the original definition of the function I gave you.



*

*How can I evaluate the indefinite integral of the above version?


One may define the "incomplete" beta function as
$$\text B_z(x,y) = \int_0^z t^{x-1}(1-t)^{y-1}dt$$
This is an antiderivative (obviously, per the fundamental theorem of calculus), but not necessarily the entire class of them. Ben Longo in the comments suggests $B_z(n,n+1) - B_z(n+1,n+1)$ as one, but I'm not sure how they arrived at this answer. I think it's safe to say, though, that any antiderivative will likely rely on more special functions if not boil down to manipulations of the above function in some way.




*How to prove the LHS - RHS equality in the aforementioned problem?


To restate the problem in language more familiar to most, $C_r  := \, _n C_r = \binom n r$ (the $n$ being omitted because it's implied I guess) and thus we seek to show
$$\sum_{k=0}^{n+1} \binom n k \frac{(-1)^k}{(n+k)(n+k+1)} = \int_0^1x^{n-1}(1-x)^{n+1}dx$$
To prove this equality, expand the parenthetical in the integral using the binomial expansion formula of $(a+b)^n = \sum_{k=0}^n \binom n k a^k b^{n-k}$:
$$\mathcal I := \int_0^1x^{n-1}(1-x)^{n+1}dx =  \int_0^1x^{n-1} \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k x^kdx$$
You can bring the $x^{n-1}$ into the summation by distributing it, and then exchange the sum and integral (with no issue since the sum is finite and the interval is finite). Then we just evaluate as normal.
$$\begin{align}
\mathcal I &= \int_0^1 \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k x^{n+k-1}dx \\
&= \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k \int_0^1 x^{n+k-1}dx \\
&= \sum_{k=0}^{n+1} \left. \binom {n+1} k \frac{ (-1)^k}{n+k} x^{n+k} \right|_{x=0}^1\\
&= \sum_{k=0}^{n+1} \binom {n+1} k \frac{ (-1)^k}{n+k}
\end{align}$$
This is similar to our intended sum, but not quite the exact same. But some manipulations can be done to it. Begin by pulling out the first term of the sum for simplicity, and then use Pascal's identity - $\binom p q + \binom p {q-1} = \binom{p+1}{q}$ - to split up the binomial coefficient. Then we can distribute and make our sum into two separate ones. This brings us to this:
$$\mathcal I = \frac 1 n + \sum_{k=1}^{n+1} \binom n k \frac{ (-1)^k}{n+k} + \sum_{k=1}^{n+1} \binom n {k-1} \frac{ (-1)^k}{n+k}$$
For now, we'll reabsorb the $1/n$ into the left-hand sum, to start it back at $k=0$. The second we'll reindex with $m=k-1$.
$$\mathcal I = \sum_{k=0}^{n+1} \binom n k \frac{ (-1)^k}{n+k} + \sum_{m=0}^{n} \binom n {m} \frac{ (-1)^{m+1}}{n+m+1}$$
Now, an interesting fact. $\binom p q$, by convention, is defined to be $0$ whenever $q>p$. For instance, $\binom 4 5 = 0$. We'll abuse that fact here, and add an additional term into the end of the right-hand sum. It'll be $0$ since the coefficient will be $\binom{n}{n+1} = 0$, but it'll make life easier later on by ensuring the bounds of the sum are the same. Similarly, for intuition, let's rename $m$ back to $k$ since it's just a dummy variable really.
$$\mathcal I = \sum_{k=0}^{n+1} \binom n k \frac{ (-1)^k}{n+k} + \sum_{k=0}^{n+1} \binom n {k} \frac{ (-1)^{k+1}}{n+k+1}$$
At this point, we use one of the factors of $-1$ from the right-hand sum to allow us to subtract that whole sum, and then combine the sums into a single one:
$$\mathcal I = \sum_{k=0}^{n+1} \binom n k (-1)^k \left( \frac{1}{n+k} - \frac{1}{n+k+1} \right)$$
Now we just combine the two fractions into one to complete the proof!
$$\mathcal I = \sum_{k=0}^{n+1} \binom n k \frac{(-1)^k }{(n+k)(n+k+1)}$$
Thus, in summary,
$$\int_0^1x^{n-1}(1-x)^{n+1}dx = \sum_{k=0}^{n+1} \binom {n+1} k \frac{ (-1)^k}{n+k} = \sum_{k=0}^{n+1} \binom n k \frac{(-1)^k }{(n+k)(n+k+1)}$$

Finally, a huge thanks to Brian M. Scott, who showed me how to prove the two sums above were equal (see here)!
