3. a) Let $n \in \mathbb{N}^{*}$. Calculate $\int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x$.

b) Calculate $\lim _{n \rightarrow \infty} \int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} x}{x^{2}+x+1} d x$.
Solution. a) $\int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x=\int_{\frac{1}{n}}^{n} \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x=\left.\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x+1}{\sqrt{3}}\right|_{\frac{1}{n}} ^{n}=\frac{2}{\sqrt{3}}\left(\operatorname{arctg} \frac{2 n+1}{\sqrt{3}}-\operatorname{arctg} \frac{\frac{2}{\sqrt{3}}+1}{\sqrt{3}}\right)$

b) Let $I_{n}=\int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} x}{x^{2}+x+1} d x$. Since $\operatorname{arctg} x+\operatorname{arctg} \frac{1}{x}=\frac{\pi}{2}, \quad \forall x>0$, we have:

![](https://cdn.mathpix.com/cropped/2024_06_07_eb436281f7e263db6a33g-2.jpg?height=214&width=1607&top_left_y=1687&top_left_x=194)

With the change of variable $t=\frac{1}{x}$ we get $\int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} \frac{1}{x}}{1+\frac{1}{x}+\frac{1}{x^{2}}} \cdot \frac{-1}{x^{2}} d x=\int_{n}^{\frac{1}{n}} \frac{\operatorname{arctg} t}{t^{2}+t+1} d t=-I_{n}$, thus

$I_{n}=\frac{\pi}{2} \int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x-I_{n}$. Considering the result from part a), we get $I_{n}=\frac{\pi}{2 \sqrt{3}}\left(\operatorname{arctg} \frac{2 n+1}{\sqrt{3}}-\operatorname{arctg} \frac{\frac{2}{n}+1}{\sqrt{3}}\right)$.

$\lim _{n \rightarrow \infty} I_{n}=\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\frac{\pi^{2}}{6 \sqrt{3}}$.

Scoring.

| a) Calculation of the integral | $2 \mathrm{p}$ |
| :--- | :--- |
| b) Calculation of $I_{n}$ | $4 \mathrm{p}$ |
| Calculation of $\lim _{n \rightarrow \infty} I_{n}$ | $1 \mathrm{p}$ |