How can I represent integral from 0 to 1 of 2/(3x^4+16) dx as the sum of an infinite series?
Thanks.
$\displaystyle \frac{2}{3x^4 + 16} =$
$\displaystyle \frac{2}{16 - (-3x^4)} =$
$\displaystyle \frac{2}{16} \cdot \frac{1}{1 - \left(-\frac{3x^4}{16}\right)} =$
$\displaystyle \frac{1}{8}\left(1 - \frac{3x^4}{2^4} + \frac{3^2 x^8}{2^8} - \frac{3^3 x^{12}}{2^{12}} + ...\right)$
I'll leave the definite integral to you ...