How do we prove that $\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7}$ converges? Tell if the following improper integral converges or not.
$$\int_0^{+\infty} \dfrac{1}{x^2+7}\ \text{d}x$$
I know that a necessary but not sufficient condition is that $\lim_{x\to +\infty} f(x) = 0$. Here this holds, so the integral might converge.
Now, I thought of the comparison:
$$\int_0^{+\infty} \dfrac{1}{x^2+7}\ \text{d}x < \int_0^{+\infty} \dfrac{1}{x^2}\ \text{d}x$$
Since the second integral diverges, and the first one is $<$, it converges.
Is this method valid?
 A: According to your efforts, I would recommend you to split the integral into two parts as follows:
\begin{align*}
\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7} & = \int_{0}^{1}\frac{\mathrm{d}x}{x^{2} + 7} + \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7}
\end{align*}
The first part is integrable because its argument is a continuous function on the compact interval $[0,1]$, hence uniformly continuous and integrable. The second part is also integrable. That is because we can apply the monotonicity of the integral to conclude that:
\begin{align*}
\int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7} < \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2}} = -\frac{1}{x}\bigg\rvert_{1}^{\infty} = 1 < + \infty
\end{align*}
Hopefully this helps!
A: In a simpler way, we can say
$$\dfrac{1}{x^2+7} < \dfrac{1}{x^2+1}$$
Thence
$$\int_0^{+\infty} \dfrac{1}{x^2+7} \text{d}x < \int_0^{+\infty} \dfrac{1}{x^2+1}\ \text{d}x$$
The second integral gives the arctangent function which evaluates to $\pi/2$. Hence
$$\int_0^{+\infty} \dfrac{1}{x^2+7} \text{d}x < \dfrac{\pi}{2}$$
Whence it converges.
To prove that the integral of $\dfrac{1}{x^2+1}$ converges indeed, see @Atila Correia's answer!
A: For $a>0$ we have  $$\int\limits_0^\infty {1\over x^2+a^2}\,dx=
{1\over a}\arctan\left ({x\over a}\right )\bigg\rvert_{0}^{\infty}={\pi\over 2a}$$
In OP case we have $a=\sqrt{7}.$
