# Converging/Diverging Integrals

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent":

$$\int_4^\infty \frac{1}{x^2+1}\,dx.$$

I am so lost. I know that it is similar to $\frac{1}{x^2}$ but how do I know if it is smaller or larger?

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There is a closed form... – Ishan Banerjee Jan 28 '13 at 16:48
Hint: ${d\over dx}\arctan x={1\over 1+x^2}$. – JohnD Jan 28 '13 at 16:51
What I did before was say that tan^-1(x) was the integral and I evaluated that between 4 and t. I got that the answer was 1.326+(pi/2) but apparently it wasn't correct! – Gabrielle Jan 28 '13 at 16:55
If you didn't know the indefinite integral, you could have used the inequality $1/(x^2+1)<1/x^2$ and compared with $\int_4^\infty 1/x^2\,dx=1/4<\infty$ to show that the integral is convergent. (This works because the integrand is positive.) – Harald Hanche-Olsen Jan 28 '13 at 17:06

The antiderivative of $\frac{1}{x^2+1}$ is $\arctan(x)$, so the integral is $\lim\limits_{x\to \infty} (\arctan(x))-\arctan(4)$. I assume you know the limit of $\arctan(x)$ as $x\to\infty$? (HINT: $\tan(x)=\frac{\sin(x)}{\cos(x)}$, what angle $x$ do you know so that this approaches $\infty$?)