Limit of definite integral with parameter How would I go around a problem like below:
$$\lim_{\alpha\to +\infty}\alpha \int_{-\alpha}^{1+\alpha}\frac{dx}{1+x^2+\alpha}$$
I believe I should deal with the denominator somehow and achieve $\arctan$ as a  primitive out of all of this. I cannot think of a solution, though.
Thank you in advance. 
 A: $$\lim_{\alpha\to +\infty}\alpha \int_{-\alpha}^{1+\alpha}\frac{dx}{1+x^2+\alpha}
=\lim_{\alpha\to +\infty}\frac{\alpha}{\sqrt{1+\alpha}} \arctan \frac{x}{\sqrt{1+\alpha}}\Bigg|_{-\alpha}^{1+\alpha}$$
$$=\lim_{\alpha\to +\infty}\frac{\alpha}{\sqrt{1+\alpha}} ( \arctan \frac{1+\alpha}{\sqrt{1+\alpha}} -\arctan \frac{-\alpha}{\sqrt{1+\alpha}})= \lim_{\alpha\to +\infty}\frac{\alpha}{\sqrt{1+\alpha}} ( \arctan \sqrt{1+\alpha} +\arctan \frac{\alpha}{\sqrt{1+\alpha}})$$
$$\lim_{\alpha\to +\infty}\frac{\alpha}{\sqrt{1+\alpha}} ( \frac{\pi}{2} +\frac{\pi}{2})=+\infty$$
A: $$\begin{align}I&=\lim_\limits{\alpha\to\infty}\alpha\int_{-\alpha}^{\alpha+1}\dfrac{\mathrm dx}{1+x^2+\alpha}\\&=\lim_\limits{\alpha\to\infty}\alpha\int_{-\alpha}^{\alpha+1}\dfrac{\mathrm dx}{(\sqrt{\alpha+1})^2+x^2}\\&=\lim_\limits{\alpha\to\infty}\alpha\cdot\dfrac1{\sqrt{\alpha+1}}\arctan\left(\dfrac x{\sqrt{\alpha+1}}\right)\Bigg|_{-\alpha}^{\alpha+1}\\&=\lim_\limits{\alpha\to\infty}\dfrac \alpha{\sqrt{\alpha+1}}\left[\arctan\left(\sqrt{\alpha+1}\right)-\arctan\left(-\dfrac{\alpha}{\sqrt{\alpha+1}}\right)\right]\\&=\lim_\limits{\alpha\to\infty}\dfrac\alpha{\sqrt{\alpha+1}}\arctan\left(\dfrac{\sqrt{\alpha+1}+\dfrac\alpha{\sqrt{\alpha+1}}}{-\alpha}\right)\\&=\lim_\limits{\alpha\to\infty}\dfrac{\alpha}{\sqrt{\alpha+1}}\arctan\left(\dfrac{2\alpha+1}{-\alpha\sqrt{\alpha+1}}\right)\\&=(+\infty)\cdot\arctan(-\infty)\\&=(+\infty)\cdot\left(\dfrac{3\pi}4\right)\\&=+\infty\end{align}$$
