Find an upper bound to $\int_0^1 \frac{x^n}{\ln (1+x)} dx$. I've conjectured that
$$\int_0^1 \frac{x^n}{\ln (1+x)} dx $$
where $n$ is a positive integer, goes to $0$ as $n$ grows larger. Knowing this, I wanted to find a non-constant function $f(n)$ such that
$$\int_0^1 \frac{x^n}{\ln (1+x)}dx \leq f(n)$$
where $f(n)$ goes to $0$ as $n$ goes to infinity. Notice that we can write the integral as
$$\lim_{\epsilon \to 0^+}\int_\epsilon^{1+\epsilon} x^n/\ln(1+x)dx $$
since it's improper.
My attempt was
$$\int_0^1\frac{x^n}{\ln (1+x)}dx \leq [\max_{x\in(0,1)} (x)]^n \int_0^1 1/\ln(1+x)dx.$$
However the max is $1^n=1$, so it does not depend on $n$. Moreover, the last integral does not converge.
Any help is appreciated.
 A: Your approach also won't work because $\int_{0}^1\frac{dx}{\log(1+x)}$ doesn't converge. $\log(1)=0$ makes the $x=0$ problematic.
Show for $x\in[0,1]$ that:
$$e^{x/2}\leq 1+x\tag1$$
The inequality follows from the power series for $e^{x/2},$ but it takes a few steps to prove.
Taking the log of $(1)$ you get:
$$\frac x2\leq \log(1+x)$$
From this, deduce a suitable $f.$
A: $\lim_{x\to0}\frac{\ln(1+x)}x=1$ hence the continuous function $x\mapsto\frac x{\ln(1+x)}$ is bounded on $(0,1]$, say by some $M>0,$ so
$$\int_0^1\frac{x^n}{\ln(1+x)}\,\mathrm dx\le M\int_0^1x^{n-1}\,\mathrm dx=\frac Mn.$$
A: You have
$${\ln(1+ x) \over x} = {\ln(1 + x) - \ln(1 + 0) \over x - 0}$$
By the mean value theorem, for each $x$ there's a $y \in [0,x]$ such that
$${\ln(1 + x) - \ln(1 + 0)  \over x - 0} = {1 \over 1 + y}$$
Since $0 \leq y \leq 1$ we therefore have
$${1 \over 2} \leq {\ln(1 + x) - \ln(1 + 0)  \over x - 0} \leq 1$$
In other words, we have
$${1 \over 2} \leq {\ln(1+ x) \over x} \leq 1$$
Equivalently,
$${x \over 2} \leq \ln(1 + x) \leq x$$
Inserting this into your integral should give you what you want.
