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I've been trying to integrate this:

$$\int_0^\infty \frac{1}{x^2 + 2x + 2} \mathrm{d} x .$$

Unfortunately I haven't found a way so far. I've been trying to factor the denominator in order to end up with partial fractions. Is there a way to factor it? If so, I can't remember any, so if you could remind me how to do it, it would be nice.

Thanks for your help.

Martin Sleziak
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Pacane
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3 Answers3

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Try using the equation: $x^2+2x+2=(x+1)^2+1$

Thomas Andrews
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  • Thanks, that would make $arctan(x+1)$ + C However, as I said, I can't remember how to factor it, could you point out how you make it? I mean, I can do it, but what are the circumstances where you have to do this? Thanks. – Pacane Apr 20 '11 at 23:35
  • @Pacane: This is a special case of "completing the square." Wikipedia has an article on it. – Cheerful Parsnip Apr 20 '11 at 23:49
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    @Pacane: Use substitution $u=x+1$ – Américo Tavares Apr 20 '11 at 23:50
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    @Pacane Well, it doesn't factor, at least over the real numbers. It does factor over the complex numbers, in which case, you can write it as: $(x+1-i)(x+1+i)$. As for factoring quadratics, see the Wikipedia article mentioned by Jim Conant: http://en.wikipedia.org/wiki/Completing_the_square – Thomas Andrews Apr 20 '11 at 23:56
  • If trying to use partial fractions, the first thing I would do is check the discriminant $b^2 - 4ac$. This will immediately tell you if it's irreducible or not, and if it is you can use the quadratic formula to find the zeroes, and hence the factorization into linear factors. – Patrick Mar 20 '12 at 15:46
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$$ \int\limits_0^\infty \frac{dx}{x^2+2x+2}= \lim\limits_{t\to\infty}\int\limits_0^t\frac{dx}{x^2+2x+2}= \lim\limits_{t\to\infty} \arctan(x+1)|_0^t= $$ $$ \lim\limits_{t\to\infty} \arctan(t+1)-\arctan(1)=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4} $$

Norbert
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KaySid
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Integrating by expanding into partial fractions nets the same result as a trigonometric substitution but with extra steps.

$$x^2+2x+2 = (x-(-1+i)) (x - (-1-i))$$

$$\begin{align*} \implies I &= \int_0^\infty \frac{dx}{x^2+2x+2} \\[1ex] &= \frac i2 \int_0^\infty \left(\frac1{x+1+i} - \frac1{x+1-i}\right) \, dx \\[1ex] &= \frac i2 \log\left(\frac{x+1+i}{x+1-i}\right) \bigg|_0^\infty \\[1ex] &= -\frac i2 \log\left(\frac{i-(x+1)}{i+(x+1)}\right) \bigg|_0^\infty \\[1ex] &= \arctan(x+1)\bigg|_0^\infty \\[1ex] &= \frac\pi4 \end{align*}$$

user170231
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