Evaluating $\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5}\mathrm{d}x$

Question

How to evaluate $$\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$

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My Attempt:

$$I = \int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$

$$I = \int_{0}^{\infty} \frac{(1-x)}{(1 - x^6)} dx$$

Substituting $t = \frac{1}{x}$

$$I = \int_{0}^{\infty} \frac{(1-\frac{1}{x})}{(1 - \frac{1}{x^6})} \cdot \frac{1}{x^2} dx$$

$$I = \int_{0}^{\infty} \frac{(1-{x})(1 + x + x^2)}{(1-x^6 )} dx$$

This does not seem to work.

Answer 1

Note that by replacing $x\to \frac{1}{x}$ $$I=\int_0^{\infty} \frac{1-x}{1-x^6}dx=\int_0^{\infty} \frac{(1-x)x^3}{1-x^6}dx$$Adding $$2I=\int_0^{\infty} \frac{(1-x)(1+x^3)}{1-x^6}= \int_0^{\infty} \frac{1-x}{1-x^3}=\int_0^{\infty} \frac{1}{x^2+x+1}$$ which should be easy..