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

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

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

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

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 $$x=1$$.

• I want to find the maxima of $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx$ not $f(x)$ which you have defined above. Sep 8, 2021 at 15:16
• Thank you for clarifying. Maxima and minima are both terms that refer to a function, hence my confusion. But the definite integral $\int_0^1 \frac{x^{n+1}}{x+1}dx$ has an exact value (dependent on $n$). When you say maxima or minima of the definite integral, I believe you mean upper bound or lower bound for the definite integral, respectively. Sep 8, 2021 at 16:10
• Yes I want to prove that $\displaystyle\frac{1}{2(n+2)}<\int\limits_0^1\frac{x^{n+1}}{x+1}dx$, I didn't know the correct term to use. Sep 8, 2021 at 16:49

Hint

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

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 $$[0,1].$$