Find maximum of $\log(1+x)(1-I_x(a,b-a))$ I am struggling to find (or at least set some bounds on) 
 $$\arg \max_x \log(1+x)(1-I_x(a,b-a)),$$
where $I_x(a,b-a)$ is the regularized incomplete beta function, i.e
$$I_x(a,b-a) = \frac {\int_0^x t^{a-1} (1-t)^{b-a-1}dt}{B(a, b-a)}$$
For $0<x<1$ with $a>1,b-a>1$.
I've noticed that this can be represented as the product of the following integrals:
$$\arg \max_x \int_0^x \frac{\mathrm{d}\tau}{1+\tau} \cdot \int_x^1 \frac{\tau^{a-1}(1-\tau)^{b-a-1}}{B(a,b-a)}\mathrm{d}\tau $$
Yet, it didn't help much to obtain the maximum nor the arg max. I've tried through the first-order derivative,  tried to bound it with arithmetic mean-geometric mean, and even tried to apply Jenssen inequality, however, without any success.
I got a hunch that since the first term does not have an explicit maximum, the only term that impacts on the maximum is the second integral, which reaches a maximum at $x\to \frac{a-1}{b-2}$, but didn't succeed to prove this bound. 
Maybe L'hoptial rule can help get some insights on this? 
Any ideas, help, and clues are much appreciated.   
 A: I found a decent approximation to my question using right Riemann_sum. 
Recalling that with Riemann sum approximation, the absolute approxination error is bounded by
$$\left\vert \int_a^b f(\tau) \mathrm{d}\tau - A_{right}\right\vert \leq \frac{M_1\cdot (b-a)}{2n},$$ 
where $M_1$ is the maximum value of the absolute value of $f^{\prime}(\tau)$ on the interval. In our case both the derivative of $\log(1+x)$ and the derivative of $(1-I_x(a,b-a))$ are moderate which is good for our purposes.
Accordingly  , we seek for
$$\arg\max_x \log(1+x)(1-I_x(a,b-a))$$
or
$$\arg \max_x \int_0^x \frac{\mathrm{d}\tau}{1+\tau} \cdot \int_x^1 \frac{\tau^{a-1}(1-\tau)^{b-a-1}}{B(a,b-a)}\mathrm{d}\tau.$$
With $n=1$, this is approximately 
\begin{align*}\arg \max_x &\frac{1}{1+x}\cdot x \cdot\frac{x^{a-1}(1-x)^{b-a-1}}{B(a,b-a)}\cdot(1-x)&\\
 = \arg\max_x&\frac{x^{a}(1-x)^{b-a}}{(1+x)B(a,b-a)}\end{align*}
This expression, however, is much easier to handle. Its derivative is
$$\frac{\partial}{\partial x } = \frac{x^{a-1} (1-x)^{-a+b-1} (-a + (-a+b+1)x +(b-1) x^2)}{B(a,b-a)(x+1)^2} $$ 
which has roots in zero, in one, and more interestingly, in 
$$x \to \frac{\pm\sqrt{a^2+2 a b-6 a+b^2+2 b+1}+a-b-1}{2 (b-1)}$$
Any further suggestions and comments are much appreciated!
A: Not an answer, I have graphed this, and it's really a tricky function. But your assumption

$x \to \frac{a-1}{b-2}$

is not true. For example, $a=0.7, b = 1$, $a=10, b=14$ Good Luck! 

A: One more thing: I tried using McLoren series for logarithm $(1+x)$ at 0 and 1. I really don't understand why both estimations go so terribly wrong. Note that I substitute $a, b-a$ for $a,b$
https://www.desmos.com/calculator/yvvenwyafg
