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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

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

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