Showing that $\int_{0}^{\infty} \cfrac{1}{x^2+t}dx$ is convergent for $t>0$ I know that $\int_{0}^{\infty} \cfrac{1}{x^2+t}dx$ is continuous on $[0,\infty)$, but I don't really know how to prove that it's convergent I thought maybe $$ \exists C:\cfrac{1}{x^2+t}\leq C\frac{1}{t}$$ seeing that $t>0$ thus: $$\int_{0}^{\infty} \cfrac{1}{x^2+t}dx\leq \int_{0}^{\infty} C\frac{1}{t}dx$$ but this doesn't really prove that it's convergent can anyone help me with this?
 A: HINT:
Write the integral of interest as
$$ \int_0^\infty \frac1{x^2+t}\,dx =\int_0^1 \frac1{x^2+t}\,dx+\int_1^\infty \frac1{x^2+t}\,dx$$
The first integral clearly converges.  For the second, note that for $t>0$,
$$0<\frac1{x^2+t}<\frac1{x^2}$$
A: We can take an antiderivative:
$$\int\frac{dx}{x^2+t} = \frac1{\sqrt{t}}\int\frac{\frac{1}{\sqrt{t}}dx}{\left(\frac{x}{\sqrt{t}}\right)^2+1}=\frac{1}{\sqrt{t}}\tan^{-1}\left(\frac{x}{\sqrt{t}}\right)$$
Now, for the definite integral, we can try to take a limit:
$$\int_0^\infty\frac{dx}{x^2+t} =\lim_{b\to\infty}\left(\frac{1}{\sqrt{t}}\tan^{-1}\left(\frac{b}{\sqrt{t}}\right) - \frac{1}{\sqrt{t}}\tan^{-1}\left(0\right)\right)=\frac{\pi}{2\sqrt{t}}$$
Since the limit exists, the improper integral is convergent.

If all you need to know is convergence, this is overkill in a way. We're looking at $x\geq 0$, and we can say $0\leq \frac{1}{x^2+t}\leq \frac{1}{x^2}$. The given integral is clearly bounded between $0$ and $1$, and if we know that $\int_1^\infty\frac{dx}{x^2}$ converges, then we can say that the given integral converges as well by a  comparison test.
A: $\int \frac{1}{t+x^2}dx=\frac{1}{\sqrt{t}}\int\frac{1}{1+(x/\sqrt{t})^2}d(x/\sqrt{t})=\frac{1}{\sqrt{t}}\arctan \frac{x}{\sqrt{t}}.$
$\lim_{x \to \infty} \arctan (x/\sqrt{t})=\frac{\pi}{2}.$
$\arctan 0= 0$
