Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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



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

share|cite|improve this answer
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. $4x^2+4x+5=4(x+?)^2+??$

share|cite|improve this answer

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

share|cite|improve this answer
  • 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: $$ \int\frac{du}{{u^2 + a^2}} = \frac{1}{a} \arctan \left(\frac{u}{a}\right)+C $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.