<s>[INST] Limit of $\int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\mathrm{d}x$ without Taylor expansion $\def\d{\mathrm{d}}$Consider $$I_n(a) = \int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\d x, \quad n\ge 1,\ a \in [0, 1)$$
and evaluate
$$\lim _{n \to \infty} I_n(a). \quad a \in (0, 1)$$
One thing I have found it that
$$\int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\d x = \int _{\frac{1}{n}}^n \frac{\arctan \frac{1}{x}}{x^2 + 2ax + 1}\,\d x$$
Also, I know $$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}. \quad \forall x > 0$$
I didn't make it further with that method, so I tried this:
$$\frac{1}{n} \le x \le n,\\
\arctan \frac{1}{n} \le \arctan x \le \arctan n.$$
So I deduced $$I_n(a) \le \frac{\pi}{2} \int _{\frac{1}{n}}^n \frac{\d x}{x^2 + 2at + 1}. \tag{1}$$
The discriminant of the denominator is $4 (a^2 - 1) \le 0\ (\forall a \in [0, 1))$, so $(1)$ becomes
$$I_n(a) \le \int _{\frac{1}{n}}^n \frac{\d x}{(x+a)^2 + \left(\sqrt{\frac{1-a^2}{a}}\right)^2}.$$
I have to get to the answer $$\frac{\pi}{4 \sqrt{1-a^2}} \arctan \frac{\sqrt{1-a^2}}{a}.$$
 [/INST] You are almost there. Since
$$
I_n(a)=\int_{1/n}^n\frac{\arctan x}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\arctan (1/x)}{x^2+2ax+1}\,dx
$$
and
$$
\arctan x+\arctan(1/x)=\frac\pi2
$$
you find that
$$
2I_n=\int_{1/n}^n\frac{\arctan x+\arctan(1/x)}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\pi/2}{x^2+2ax+1}\,dx.
$$
Hence
$$
I_n(a)=\frac\pi4\int_{1/n}^n\frac{1}{x^2+2ax+1}\,dx
=\frac{\pi}{4}\Bigl[\frac{1}{\sqrt{1-a^2}}\arctan\frac{a+x}{\sqrt{1-a^2}}\Bigr]_{1/n}^{n}.
$$
Insert the limits, take the limit $n\to+\infty$ and finally use $\arctan x+\arctan(1/x)=\pi/2$ again, and you are done.
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