Prove that I(n) is a strictly decreasing series: I(n)= $\int_0^1 \frac{(x^n)}{(x+1)} dx$ Prove that $I(n)$ is a strictly decreasing series:
$$I(n)=  \int_0^1 \frac{(x^n)}{(x+1)} dx.$$
If it were just decreasing I would have used the fact that when $x$ is between $0$ and $1$ ($0$ and $1$ are included), $x^n \geq x^{n+1}$. Then divide by $x+1$ and integrate. But I found the version of the same problem where it states that it has to be strictly decreasing. Any ideas?

Hint.
A continuous non-negative function $f$ having a strictly positive value has a stricly positive integral.
You can apply this to the function $g(x)=\frac{x^n}{x+1}-\frac{x^{n+1}}{x+1}$.
