What is the value of $\int_{0}^{t} (t-s)^{\alpha-1} (1-s)^{\alpha} \rm{d}s$?, where $0<t<1$ and $0 <\alpha \leq 1$.

Question

I saw the value of this integral in one research article. They wrote $-\frac{1}{\Gamma{(\alpha+1)}\Gamma{(\alpha)}}\displaystyle\int_{0}^{t} (t-s)^{\alpha-1} (1-s)^{\alpha} \rm{d}s = -\frac{t^{\alpha}}{\Gamma(\alpha+1)} + \frac{t^{\alpha+1}}{\Gamma(\alpha+2)}$. I don't know how they got this value for that integral value. I know how to get value for this type of integral $\Rightarrow \frac{1}{\Gamma{(\alpha+1)}\Gamma{(\alpha)}}\displaystyle\int_{0}^{t} (t-s)^{\alpha-1} s^{\alpha} \rm{d}s = \frac{t^{2\alpha}}{\Gamma(2\alpha+1)}$.
Let $u=s/t$. Then $\frac{1}{\Gamma{(\alpha+1)}\Gamma{(\alpha)}}\displaystyle\int_{0}^{t} (t-s)^{\alpha-1} s^{\alpha} \rm{d}s = t^{2\alpha} \int_{0}^{1} (1-u)^{\alpha-1}u^{\alpha}\rm{d}u = t^{2\alpha} \beta(\alpha+1,\alpha) = \frac{t^{2\alpha}}{\Gamma(2\alpha+1)}$. But how to solve $\displaystyle\int_{0}^{t} (t-s)^{\alpha-1} (1-s)^{\alpha} \rm{d}s$?