<s>[INST] Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx\right)<\frac{1}{2(n+1)(n+2)}$$
We will be able to prove the above inequality
 [/INST] Simply observe that for $x \in [0,1]$ and $n \in \mathbb Z^+$,
$$0 \le (1-x)x^n = (x+1)x^n - 2x^{n+1},$$ the first inequality arising from the fact that $x \ge 0$ and $1-x \ge 0$.  Consequently,
$$0 \le \frac{x^{n+1}}{x+1} \le \frac{x^n}{2},$$ and integrating gives the desired inequality.
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