# How to evaluate $\int^\infty_{-\infty} \frac{dx}{4x^2+4x+5}$?

I need help in my calculus homework guys, I can't find a way to integrate this, I tried use partial fractions or u-substitutions but it didn't work.

$$\int^\infty_{-\infty} \frac{dx}{4x^2+4x+5}$$

Thanks much for the help!

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Please, make titles informative. Note what I changed the question title to, so you can think of something similar next time. – Pedro Tamaroff Mar 18 '13 at 22:46

Hints:

$$(1)\;\;\;\;\int\limits_{-\infty}^\infty\frac{dx}{4x^2+4x+5}=\int\limits_{-\infty}^\infty\frac{dx}{4\left(x+\frac{1}{2}\right)^2+4}=\frac{1}{4}\int\limits_{-\infty}^\infty\frac{dx}{1+\left(x+\frac{1}{2}\right)^2}$$

$$(2)\;\;\;\text{ For a derivable function}\;f(x)\;,\;\;\int\frac{f'(x)}{1+f(x)^2}dx=\arctan(f(x))+C\ldots$$

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I think in English the term is "differentiable". (vs. "derivable" in Spanish) – Pedro Tamaroff Mar 18 '13 at 22:46
¡ Qué sé yo...! También existe derivable pero quizás tenga un significado algo distinto pues "differentiable", ahora que me recuerdas, sí lo he visto más seguido. Gracias. – DonAntonio Mar 18 '13 at 22:47
$\displaystyle \frac{1}{4}\int\limits_{-\infty}^\infty\frac{dx}{1+\left(x+\frac{1}{2}\right)^2}‌​$, perhaps? – J.H. Mar 18 '13 at 22:52
Thanks, @J.H.. I missed that first $\,4\,$ in the denominator. I shall edit – DonAntonio Mar 18 '13 at 22:54

Try completing the square in the denominator, i.e. $4x^2+4x+5=4(x+?)^2+??$

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Use Residue Theorem. $$\int_{-\infty}^{\infty}\frac{dx}{4x^2+4x+5}=\int_C\frac{dz}{4z^2+4z+5}=2\pi i\text{Res}|_{z=-\frac{1}{2}+i}=2\pi i\frac{1}{2i}=\pi.$$

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• Manipulate the denominator to get $(2x+1)^2 + 4 = (2x+1)^2 + 2^2$.

• Let $u = 2x+1 \implies du = 2 dx \implies dx = \frac 12 du$,

• $\displaystyle \frac 12 \int_{-\infty}^\infty \dfrac{du}{u^2 + (2)^2}$

• use an appropriate trig substitution which you should recognize: $$\int\frac{du}{{u^2 + a^2}} = \frac{1}{a} \arctan \left(\frac{u}{a}\right)+C$$

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