Use a power series to approximate the definite integral, $I$, to six decimal places. \[ \begin{array}{l} \int_{0}^{0.3} \frac{1}{1+x^{6}} d x \\ \square \end{array} \]

Step 1 ：We are given the integral \(\int_{0}^{0.3} \frac{1}{1+x^{6}} dx\) and we are asked to approximate it to six decimal places using a power series.

Step 2 ：The power series for \(\frac{1}{1+x}\) is \(1 - x + x^2 - x^3 + x^4 - x^5 + \cdots\).

Step 3 ：We can use this power series to approximate the integral of \(\frac{1}{1+x^6}\) from 0 to 0.3 by substituting \(x^6\) for \(x\) in the power series.

Step 4 ：We then integrate term by term from 0 to 0.3.

Step 5 ：We add up the terms until the absolute value of the last term is less than \(10^{-6}\), which will give us an approximation to six decimal places.

Step 6 ：The definite integral, \(I\), to six decimal places is approximately \(\boxed{0.299969}\).