Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$

Question

$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$

$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$

If we can prove that $\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$ is always greater than $0$ we can find the minima of the function. How can we prove that the term is always positive?

And how can we prove that $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$

Answer 1

If the function you are integrating is always positive, then the integral must be positive as well. Take a look at the function being integrated. We have $$f(x) = \frac{x^{n+2}}{(x+1)^2(n+2)}$$ with $0\leq x \leq 1$. Given that $x\in [0,1]$, is it true that $f(x)$ is always positive?

I assume also that you mean to find the maxima of the $f(x)$ as above. To do this, we can rely on the fact that $f$ is a differentiable function. Hence we can use the first derivative test to find the local maxima, then compare the values of the local maxima to the endpoints $x=0$ and $x=1$.

Answer 2

Hint

You know that $x \in \left[0;1\right]$ so $$ \frac{1}{x+1}>\frac{1}{2} $$ What does this imply for $\displaystyle \int_{0}^{1}\frac{x^n}{1+x}$ ?

Answer 3

Hint $:$ For maxima observe that $x \mapsto \frac {x} {x+1}$ is increasing on $[0,1].$ For minima observe that $x \mapsto \frac {1} {x+1}$ is decreasing on $[0,1].$