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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.

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  • $\begingroup$ Think about the graph of the arctan function. $\endgroup$ – David H Apr 5 '14 at 21:25
  • $\begingroup$ Hint : $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$. $\endgroup$ – Amateur Apr 5 '14 at 21:26
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The tangent function means opposite over adjacent. The arc-tangent function takes this ratio and gives the angle that gives this ratio. So what does the triangle look like when the opposite is infinitely bigger than the adjacent?

enter image description here

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).

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Arctangent approaches a limit value:

enter image description here

This is what you need to use in the limit of integration. Indeed: $$ \int_4^\infty \frac{dx}{x^2+4^2} = \lim_{t \to \infty} \int_4^t \frac{dx}{x^2+4^2} = \lim_{t \to \infty} \left[\frac{1}{4}\arctan\frac{x}{4}\right]_{x=4}^{x=t} $$

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