Recursive relationship for incomplete beta function Consider the incomplete Beta function $I_x(a, b)$
$$ I_x(a, b) =
\dfrac{B(x; a, b)}{B(a, b)} = 
\dfrac{\int_0^x t^{a-1} \left( 1-t \right)^{b-1} dt}{\int_0^1 t^{a-1} \left( 1-t \right)^{b-1} dt}.
$$
How can I prove the recursive property given in Wikipedia
$$
I_x(a+1, b) = I_x(a, b) - \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)} 
$$

Using the Beta function property $B(a+1,b)=aB(a.b)/(a+b)$ compute the derivative
$$I_x'(a+1, b) = \frac{x^a (1-x)^{b-1}}{B(a+1,b)} = 
\frac{x^a (1-x)^{b-1}}{B(a,b)a/(b+a)}=
(a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}$$
Now let $$
f(a,b,x) = I_x(a+1, b) - I_x(a, b) + \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)} 
$$
Then $f(a,b,0)=0$ and
$$f'(a,b,x) = I_x'(a+1, b) - I_x'(a, b) + \frac{d}{dx}\Big(\frac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)}\Big)$$ 
$$f'(a,b,x)= (a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}
- \frac{x^{a-1} (1-x)^{b-1}}{B(a,b)}
+ \frac{x^{a-1}(1-x)^b}{B(a,b)}
- \frac{b x^a(1-x)^{b-1}}{aB(a,b)}$$
$$f'(a,b,x)= 
\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}
\Big((a+b)x/a - 1 + 1 -x -bx/a\Big) = 0
$$
therefore $f \equiv 0$ and the recursive property is proven.
