<s>[INST] A limit question with a parameter $\displaystyle\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx$
How to solve it?
Can I solve it in this way?
\begin{align}
&\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \lim_{a\to0} \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \frac{1}{1+x^2}\,dx\\
&=π/4
\end{align}
The answer is right, but I don't know whether the argument is correct.
 [/INST] $$1+a^2+x^2=(1+a^2)\left(1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2\right)\implies$$
$$\int_0^{1+a}\frac{dx}{1+a^2+x^2}=\frac1{\sqrt{1+a^2}}\int_0^{1+a}\frac{\left(\frac1{\sqrt{1+a^2}}\right)dx}{1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2}=$$
$$=\left.\frac1{\sqrt{1+a^2}}\arctan\frac x{\sqrt{1+a^2}}\right|_0^{1+a}=\frac1{\sqrt{1+a^2}}\arctan\frac{1+a}{\sqrt{1+a^2}}$$
Thus, in the limit you get $\;\dfrac\pi4\;$
</s>