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

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.

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

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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. – Grumpy Parsnip Apr 20 '11 at 23:49
@Pacane: Use substitution $u=x+1$ – Américo Tavares Apr 20 '11 at 23:50
@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: en.wikipedia.org/wiki/Completing_the_square – Thomas Andrews Apr 20 '11 at 23:56
Thanks for your help! – Pacane Apr 21 '11 at 0:29

$$\int\limits_0^\infty \frac{dx}{x^2+2x+2}= \lim\limits_{t\to\infty}\int\limits_0^t\frac{dx}{x^2+2x+2}= \lim\limits_{t\to\infty} \arctan(x+1)|_0^t=$$ $$\lim\limits_{t\to\infty} \arctan(t+1)-\arctan(1)=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$

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hi i am new to this forum. can anyone tell me how to compose the mathematical equations on this forum. – KaySid Dec 26 '11 at 21:37
do it like this $\int \mathrm{d}x$ (becomes $\int \mathrm{d}x$) and right click on equations to see how the source. – user16697 Dec 26 '11 at 21:39