And another real integral to be solved by contour integration I want to solve
$$\int_0^\infty\frac{1}{x^3+x^2+x+1}dx$$
and i have really learned a lot already by failing to solve it.
I want to solve it using a clever contour. It is possible to do it using partial fractions (gives me $\frac{\pi}{4}$), but i´d rather not, since the residues seem quite nice.
Obviously, poles are at $-1,i,-i$, with residues being $\frac{1}{2},-\frac{1}{4}+\frac{i}{4},-\frac{1}{4}-\frac{i}{4}$.
Here´s what ive tried:
I cant get it to be from $-\infty$ to $\infty$ since its not even or anything.
I tried a contour that looks like about a big slice of a pie, but theres no angle for the "way back in" that gives back the original equation (i hope you know what i mean).
I know a theorem that would help me if it wasnt for the real root, so that doesnt help either.
Is there really no clever contour or do i fail to see it...
 A: Hint :
Integrate the following function
$$F(z) = \frac{\log(z)}{z^3+z^2+z+1}$$
Around a key-hole contour indented at the origin.

ADDENDUM
The logarithm will cancel when you perform the paramatrization. Since we are assuming that the logarithm will be analytic on that contour we have to use the branch $(0,2\pi]$. So along the positive x-axis we will have the following
$$\int^\infty_0 \frac{\ln(x)}{x^3+x^2+x+1}dx-\int^\infty_0\frac{\ln(x)+2\pi \, i}{x^3+x^2+x+1}dx = -2\pi\, i\int^\infty_0\frac{dx}{x^3+x^2+x+1}$$
Now, assuming that the integral will vanish at the small and big circles  we have
$$-2\pi\, i\int^\infty_0\frac{dx}{x^3+x^2+x+1} = 2\pi  \,i\,\text{Res}(F(z),-1,\pm i)$$
A: It's really not easy to pick a right contour for this one since the integrand is not even. But for this very problem it's possible to reduce the task to integrating even function. It can be done the following way: after substitution $x=\frac{1}{t}$ we get
$$I=\int\limits_0^{+\infty}\frac{dx}{x^3+x^2+x+1}=\int\limits_0^{+\infty}\frac{t\,dt}{t^3+t^2+t+1},$$
$$I=\frac12\int\limits_0^{+\infty}\frac{(x+1)dx}{x^3+x^2+x+1}=\frac12\int\limits_0^{+\infty}\frac{dx}{x^2+1},$$
and the integrand is even now, so we can use standard techniques (adding to $[0;R]$ the segment $[-R;0]$, closing the contour with semicircle, etc.).
