Calculating integral $\int_{0}^{1}x^{k+1}\cdot (1-x)^l\cdot dx$

Question

I need to calculate following integral $\int_{0}^{1}x^{k+1}\cdot (1-x)^l\cdot dx$, where $k, n \in \mathbb{N}_{0}$. I tried to use Newton's binomial formula for $(1 - x)^l$ and got $\sum_{r=0}^{l} {l \choose r}\cdot (-1)^r\cdot \cfrac{1}{k+r+2}$. Maybe there is another way to do this integral, as sum seems not very good answer.

Answer 1

$$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ and so: $$\int_0^1x^{k+1}(1-x)^ldx=B(k+2,l+1)=\frac{\Gamma(k+2)\Gamma(l+1)}{\Gamma(k+l+3)}=\frac{(k+1)!(l)!}{(k+l+1)!}$$ The final part is true for $k,l\in Z$