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Please help me in this integral:

$$\lim_{R \to \infty} \int_0^R \frac{dx}{x^2+x+2}$$ I've tried as usually, but it seems tricky. Do You have an idea?

Thanks in advance!

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I've tried as usually Could you be more specific? –  Did May 15 '13 at 21:30
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2 Answers 2

up vote 2 down vote accepted

We can rewrite the integrand as

$$\frac{1}{\frac{7}{4}+(x+\frac{1}{2})^2}$$

so that the integral becomes

$$\int_{0}^{\infty}\frac{1}{\frac{7}{4}+(x+\frac{1}{2})^2}=\frac{2}{\sqrt7}\tan^{-1}\left(\frac{2x+1}{\sqrt{7}}\right)\Big|_0^{\infty}=\frac{1}{\sqrt{7}}\left(\pi-2\tan^{-1}\left(\frac{1}{\sqrt{7}}\right)\right)$$

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$$\dfrac1{x^2+x+2} = \dfrac1{\left(x+\dfrac12 \right)^2 + \left(\dfrac{\sqrt{7}}2 \right)^2}$$ Recall that $$\int_a^b \dfrac{dx}{(x+c)^2 + d^2} = \dfrac1d \left.\left(\arctan\left(\dfrac{x+c}d\right)\right)\right \vert_{a}^b$$ I trust you can finish it from here. You will also need to use the fact that $$\lim_{y \to \infty} \arctan(y) = \dfrac{\pi}2$$

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