Integrals with Taylor expansion? Can I use the Taylor series to find this integrals?
$$ \lim_{n\to \infty } \int _{0}^n \frac{\arctan x}{x^2 + x + 1}dx $$
$$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$
 A: Here is a relatively simple standard approach to evaluating the first of your integrals.
Let
$$I = \int_0^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$
Dividing up the interval of integration as follows
$$I = \int_0^1  \frac{\arctan x}{x^2 + x + 1} \, dx + \int_1^\infty  \frac{\arctan x}{x^2 + x + 1} \, dx.$$
In the second of the integrals appearing to the right, enforcing a substitution of $x \mapsto 1/x$, one has
\begin{align}
I &= \int_0^1  \frac{\arctan x}{x^2 + x + 1} \, dx + \int_0^1  \frac{\arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\
&= \int_0^1 \frac{\arctan (x) + \arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\
&= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 + x + 1} \tag1\\
&= \frac{\pi}{2} \int_0^1 \frac{dx}{\left (x + \frac{1}{2} \right )^2 + \frac{3}{4}}\\ 
&= \frac{\pi}{\sqrt{3}} \left [\arctan \left (\frac{2x + 1}{\sqrt{3}} \right ) \right ]_0^1\\
&= \frac{\pi^2}{6 \sqrt{3}}
\end{align}
At step (1) the following well-known result for the inverse tangent function has been used:
$$\arctan (x) + \arctan \left (\frac{1}{x} \right ) = \frac{\pi}{2}, \qquad x > 0.$$
