Riemann integral problem 4 How can I evaluate the following limit with Riemann integral not with telescoping series:
$$ u_n=\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right),$$
$$\lim _{n\to \infty }u_n=\lim _{n\to \infty }\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right)=\lim_{n\to \infty\ }\frac{1}{n}\sum _{k=1}^nf\left(\frac{k}{n}\right)=\int_0^1\:f\left(x\right)dx$$

Integration by parts gives:
$$\int_{0}^{1}\arctan\left(\frac{1}{1+x+x^2}\right)\,dx = \arctan\frac{1}{3}+\int_{0}^{1}\frac{x(1+2x)}{(1+x^2)(1+(x+1)^2)}\,dx$$
and the last integral equals:
$$ \int_{0}^{1}\left(\frac{x}{1+x^2}-\frac{x+1}{1+(x+1)^2}+\frac{1}{1+(x+1)^2}\right)\,dx =\frac{2\log 2-\log 5}{2}+\arctan 2-\frac{\pi}{4}$$
so, in the most compact form:
$$\int_{0}^{1}\arctan\left(\frac{1}{1+x+x^2}\right)\,dx = \arctan\frac{3}{4}+\log\frac{2}{\sqrt{5}}.$$
Don't you think something went wrong when converting your partial sums into Riemann sums?
