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.
 A: $$1+a^2+x^2=(1+a^2)\left(1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2\right)\implies$$
$$\int_0^{1+a}\frac{dx}{1+a^2+x^2}=\frac1{\sqrt{1+a^2}}\int_0^{1+a}\frac{\left(\frac1{\sqrt{1+a^2}}\right)dx}{1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2}=$$
$$=\left.\frac1{\sqrt{1+a^2}}\arctan\frac x{\sqrt{1+a^2}}\right|_0^{1+a}=\frac1{\sqrt{1+a^2}}\arctan\frac{1+a}{\sqrt{1+a^2}}$$
Thus, in the limit you get $\;\dfrac\pi4\;$
A: The parameter $a$ appears both in the integration range and in the integrand function, hence we cannot put $a=0$ for first in the integration range, then in the integrand function, at least in principle. In this case we are allowed to do so since for any $a>0$:
$$ \int_{1}^{1+a}\frac{dx}{1+x^2+a^2}\leq\int_{1}^{1+a}\frac{dx}{1+x^2}\leq\frac{a}{2}\xrightarrow[a\to 0^+]{}0 $$
and:
$$ \int_{0}^{1}\left(\frac{1}{1+x^2}-\frac{1}{1+x^2+a^2}\right)\,dx \leq a^2\int_{0}^{1}\frac{dx}{(1+x^2)^2}\xrightarrow[a\to 0^+]{}0$$
so your method is right, just needs a proper explanation.
A: The value before taking limits is: $$\frac{1}{\sqrt{1+a^2}}\cdot arctan(\frac{x}{\sqrt{1+a^2}})$$Taking limit as a goes to 0  give $\pi/4$.
