<s>[INST] 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. 
 [/INST] $$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$
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 $n=1$:
$$ \sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$
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