"Fast" way to calculate $\int_{-1}^1\dfrac{x}{x^6+\mathrm{i}} dx$ by hand?

Question

Is there a faster way to calculate

$\int_{-1}^1\dfrac{x}{x^6+\mathrm{i}} dx$

than here:

https://www.integral-calculator.com/#expr=x%2F%28x%5E6%2Bi%29&lbound=-1&ubound=1

Answer 1

Since the integrand is an odd function, i.e. $f(-x)=-f(x)$ between oposing limits, we know that the integral is zero. Proof:

$$I=\int_{-a}^a f(x)\,\textrm{d}x = \int_{-a}^0f(x)\,\textrm{d}x+\int_{0}^af(x)\,\textrm{d}x$$

Now say for the first integral $x=-z$, hence $\textrm{d}x=-\textrm{d}z$, thus

$$I=\int_{a}^0f(z)\,\textrm{d}z+\int_{0}^af(x)\,\textrm{d}x=-\int_{0}^af(z)\,\textrm{d}z+\int_{0}^af(x)\,\textrm{d}x=0$$

Answer 2

The solution is suggested in comments. Summarizing:

The function is continuous on $[-1,1]$, hence integrable. We choose a uniform division of $[0,1]$ (and symmetrically $[-1,0]$) and values in the right (symmetrically: left on $[-1,0]$) ends. Every such a sum is equal to zero, hence the integral is equal to zero.

All is done in mind, so it seems to be the quickest method.

Answer 3

$$\frac x{x^6+i}=\frac{x(x^6-i)}{x^{12}+1}$$ and both the real and imaginary parts are odd functions, hence $0$.