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

Inspired by the other answers: $f(x) = \frac{x}{1+x}$ is concave on $[0, \infty)$, and therefore
$$
 f(x) \le f(1) + (x - 1) f'(1) = \frac 1 4(1+x) 
$$
for $x \ge 0$. It follows that
$$
 \int_0^1 \frac{x^n}{1+x}\, dx \le \frac 1 4 \int_0^1 \left(x^{n+1} + x^{n+2} \right) \, dx \\
= \frac 1 4 \left( \frac 1{n+2} + \frac 1{n+3}\right) < \frac{1}{2(n+2)} \, ,
$$
which is  slightly sharper than the desired estimate.
