How to show that $\int_0^x \left(\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}\right)(x-t)^n\mathrm dt\to0$ when $n\to\infty$, for $\vert x\vert < 1$? I want to show that :
$$ \forall \vert x\vert < 1 \:,\: \int_0^x \left(\dfrac{1}{(t-1)^{n+1}}-\dfrac{1}{(t+1)^{n+1}}\right)(x-t)^n\mathrm dt \xrightarrow[n\to\infty]{} 0 $$
I've tried to bound from above $(x-t) ^n$ but 
$\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}$ seems to tend towards $\infty$.
I've also tried with absolute value (hope to find an upper bound) but my attemps failed. I don't see how to show that this integral tends to $0$.
 A: Split the integral into the two parts
$$
\int_0^x \left(\frac{x-t}{t-1}\right)^n (t-1)^{-1}\,dt - \int_0^x \left(\frac{x-t}{t+1}\right)^n (t+1)^{-1}\,dt =: I_1(n,x) - I_2(n,x).
$$
For simplicity we'll suppose that $x > 0$ and leave the $x<0$ case out of this answer. So for us $0 < x < 1$.
We can rewrite
$$
I_1(n,x) = (-1)^{n+1} \int_0^x \left(\frac{x-t}{1-t}\right)^n (1-t)^{-1}\,dt.
$$
Now
$$
\frac{d}{dt} \frac{x-t}{1-t} = \frac{x-1}{(1-t)^2} < 0
$$
for $0 \leq t \leq x < 1$, so
$$
0 \leq \frac{x-t}{1-t} \leq \left.\frac{x-t}{1-t}\right|_{t=0} = x,
$$
and thus
$$
0 \leq \int_0^x \left(\frac{x-t}{1-t}\right)^n (1-t)^{-1}\,dt \leq x^n \int_0^x (1-t)^{-1}\,dt = -x^n \log(1-x). 
$$
It follows that
$$
|I_1(n,x)| \leq -x^n \log(1-x). \tag{$*$}
$$
A similar argument can be used for $I_2(n,x)$. Since
$$
\frac{d}{dt} \frac{x-t}{t+1} = - \frac{x+1}{(t+1)^2} < 0
$$
we have
$$
0 \leq \frac{x-t}{t+1} \leq x,
$$
and thus
$$
0 \leq I_2(n,x) \leq x^n \log(1+x). \tag{$**$}
$$
Combining $(*)$ and $(**)$ yields
$$
\begin{align*}
\left| \int_0^x \left(\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}\right)(x-t)^n\,dt \right| &\leq x^n [\log(1+x)-\log(1-x)] \\
&= x^n \log\!\left(\frac{1+x}{1-x}\right). \tag{$***$}
\end{align*}
$$
Thus your integral tends to $0$ as $n \to \infty$.
A: Here is a quick and dirty way of getting the answer. Everything I am sure can be rigorously justified.
It is well known that (I believe as the Cauchy repeated integral formula)
$$\int_0^x f(t)(x-t)^n dt = I_0^n f(x)= \int_0^x \int_0^x \cdots \int_0^x f(t) dt,$$
where if you excuse the sloppiness, we have repeated integration with the same lower bound of 0.
EDIT: Not confident in my calculations below. If done right it should work. I believe the second term is wrong for each integration but they should both be equal.
The $n$th integral of $1/(x-1)^{n+1}$ is $(-1)^n/n! \cdot 1/(x-1)+(-1)^{n+1}$.
Similarly, for $1/(x+1)^{n+1}$ it is $(-1)^n/n! \cdot 1/(x+1) + (-1)^{n+1}$.
Then the limit is easily seen to be zero.
