# 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 $\infty$?)

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I'm pretty sure the limit is pi/2. This helped a lot thank you! –  Gabrielle Jan 28 '13 at 16:59
@user59974 Yes that's right, no problem. I think your answer in the comments is correct, maybe your rounding wasn't consistent with the rounding in the solution (was it something you had to fill in on the computer?). If this was helpful, you could consider accepting it as answer. –  user50407 Jan 28 '13 at 17:01
Okay I did that. Thanks again! –  Gabrielle Jan 28 '13 at 17:07