# Help in evaluating an Integral over an interval

So I have been given an Integral and its answer.

$$\int_4^\infty\frac{1}{x^2+16}\,\text{d}x$$

The book used Trig substitution and got the answer:

$${1\over 16}\left.\left(4\arctan\left({x\over 4}\right)\right)\right|_4^\infty$$*the last symbol means from $(4, \infty)$.

I know how to evaluate at 4, but I am having trouble finding out the integral at infinity. I am confused on how to evaluate and solve this when arctan goes to infinity? Please help.

• Think about the graph of the arctan function. – David H Apr 5 '14 at 21:25
• Hint : $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$. – Amateur Apr 5 '14 at 21:26

Now just keep going, what angle is $\theta$ approaching?
Hopefully you see that it approaches $90^\circ = \frac{\pi}{2}$ (you should always use radians when taking integrals involving angles).
This is what you need to use in the limit of integration. Indeed: $$\int_4^\infty \frac{dx}{x^2+4^2} = \lim_{t \to \infty} \int_4^t \frac{dx}{x^2+4^2} = \lim_{t \to \infty} \left[\frac{1}{4}\arctan\frac{x}{4}\right]_{x=4}^{x=t}$$