Inequality very difficult to show

Question

1) $\int _0^1\:\frac{x^n}{x^n+1}dx\ge \int _0^1\:\frac{x^{n+1}}{x^{n+1}+1}dx$ but I dont want to use $I_{n+1}-I_n$ 2) How we can prove with direct comparison test for ( Improper ) Integrals that is bounded: $I_n\:=\int _n^{n+1}\:\frac{2x-1}{x}dx$

P.S.: Have any idea how we can solve the second problem?

Answer 1

$$I_n:=\int_0^1\frac{x^n}{x^n+1}dx=\int_0^1\left(1-\frac1{x^n+1}\right)dx=1-\int_0^1\frac{dx}{x^n+1}$$

$$I_{n+1}=\int_0^1\frac{x^{n+1}}{x^{n+1}+1}dx=\int_0^1\left(1-\frac1{x^{n+1}+1}\right)dx=1-\int_0^1\frac{dx}{x^{n+1}+1}$$

Now, it is trivial that for $\;x\in[0,1]\;$ we have

$$\frac1{x^{n+1}+1}\ge\frac1{x^n+1}\ge0$$

and both functions are non-negative in the unit interval, and from here that $\;I_n\ge I_{n+1}\;$ .

As for (2): observe that

$$I_n=\int_n^{n+1}\left(2-\frac1x\right)\le\int_n^{n+1}2\,dx=2$$

By the way, this is not an improper integral.

Answer 2

Hints:

1)We have $x\in [0,1]\Rightarrow x^{n+1}\le x^n$. Then prove that $\frac{x^n}{x^n+1}\ge\frac{x^{n+1}}{x^{n+1}+1}$. You can use the fact that for any real number $a\neq -1$ we have $\frac{a}{a+1}=1-\frac{1}{a+1}$.

2)For any $x>0$ we have $2x-1<2x$