Use a power series to approximate the definite integral to six decimal places. Use a power series to approximate the definite integral to six decimal places. 
$\int_{0}^{0.3} \frac{x^2}{1+x^4}$
I'm not sure how to find the sum of this for solving when it has x's in the numerator, this is what I assumed however. 
$\frac{1}{1-x} $ $\sum_{n=0}^\infty $ x^n 
$\sum_{n=0}^\infty  \frac{1}{1-x^4}$ 
$\sum_{n=0}^\infty {(-x^4)}^n$ ==> $\frac{x}{1+x^4}$
I'm not sure if this is right nor what the next step is to get the approximate definite integral. 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{x^{2} \over 1 + x^{4}} & = x^{2} - x^{6} + x^{10} - x^{14} + \mrm{O}\pars{x^{16}}
\end{align}
You must truncate the series whenever the 'last term' is smaller than the Machine Precision $\ds{\texttt{mp}}$. Namely,
$$
x^{n} < \texttt{mp} \implies n < {\ln\pars{\texttt{mp}} \over \ln\pars{x}} =
{\verts{\ln\pars{\texttt{mp}}} \over \verts{\ln\pars{x}}} <
{\verts{\ln\pars{\texttt{mp}}} \over \verts{\ln\pars{0.3}}} 
$$
Typical $\ds{\texttt{mp}}$ are of order $\ds{10^{-16}}$ which truncates the series up to $\ds{x^{14}}$:
$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{0.3}{x^{2} \over 1 + x^{4}}\,\dd x} & \ds{=} & \ds{0.00896891723506713\ldots}
\\[1mm]
\ds{\int_{0}^{0.3}\pars{x^{2} - x^{6} + x^{10} - x^{14}}\,\dd x} & \ds{=} & \ds{0.0089689172\color{#f00}{289906155844}\ldots}
\end{array}\right.
$$

That's the best you can do when $\ds{\texttt{mp} \sim 10^{-16}}$.

A: $$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$
and the last series is a series with positive terms. Since
$$ \sum_{n\geq 2}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)<\frac{1}{19}\sum_{n\geq 2}\left(\frac{3}{10}\right)^{8n+3}<10^{-11} \tag{2}$$
the first eight figures of $I$ are given by the sum appearing in the RHS of $(1)$ restricted to $n=0$ and $n=1$:
$$ \sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$
