How to solve this challenging integral? How do you integrate $\int_0^1 \frac{(1-x)^{a+2}x^{b+1}}{a+2}dx$, where $a,b$ are constants? I've tried integration by parts, u-sub, and I have had no luck so far. The answer is supposed to be $\frac{(a+1)!(b+1)!}{(a+b+4)!}$.
 A: By using the Euler Beta integral, defined as 
$$\mathcal{B}(A, B) = \int_0^1 x^{A-1} (1-x)^{B-1}\ dt$$
Setting $A-1 = a+2\ $ and $\ B-1 = b+1\ $, you can easily get the result which is:
$$\frac{1}{a+2}\frac{\Gamma (a+3) \Gamma (b+2)}{\Gamma (a+b+5)}$$
The result in terms of Euler Gamma function is obtained by making use of the general relation:
$$\mathcal{B}(A, B) = \frac{\Gamma(A)\Gamma(A)}{\Gamma(A+A)}$$
To get your final result, remember that Euler Gamma is such that for $A, B \in \mathbb{N}$:
$$\Gamma(A) = (A-1)!$$
Indeed you have
$$\Gamma(A)\Gamma(B) = (A-1)!(B-1)!$$
$$\Gamma(A + B) = (A + B - 1)!$$
In your case then, expanding the whole:
$$\frac{1}{a-2}\frac{(a-2)!(b-1)!}{(a + b + 5 - 1)!} = \frac{(a-1)!(b-1)!}{(a + b + 4)!}$$
A: Binomial Method:
Call $a+2 = N$, $b+1 = M$ then:
$$\frac{1}{a+2}\int_0^1 (1+x)^N x^M\ dx$$
Expand the first term in the binomial way:
$$\frac{1}{a+2}\int_0^1 \sum_{k =0}^N \binom{N}{k} x^{N-k} x^M\ dx$$
$$\frac{1}{a+2}\sum_{k =0}^N \binom{N}{k}\int_0^1 x^{N-k+M}\ dx$$
$$\frac{1}{a+2}\sum_{k =0}^N \binom{N}{k}\frac{1}{N-k+M+1}$$
The sum is well known even if it expressed in terms of the Hypergeometric function, and the result is
$$\frac{1}{a+2}\frac{\, _2F_1(-M-n-1,-n;-M-n;-1)}{M+n+1}$$
Where $M = b+1$ and $N = a+2$. Now substitute back.
The above series can be rewritten in terms of the Beta function here below
In any case, we end up with that. 
