Answer
$0.008969$
Work Step by Step
$\int_{0}^{0.3}\frac{x^{2}}{1+x^{4}}dx=\int_{0}^{0.3}\frac{x^{2}}{1-(-x^{4})}dx$
$=\int_{0}^{0.3}x^{2}\Sigma_{0}^{\infty}(-1)^{n}(x^{4n})dx$
$=\frac{(0.3)^{3}}{3}-\frac{(0.3)^{7}}{7}+...$
$=0.008969$
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