# Thread: find maclaurin series

1. ## find maclaurin series

integral of 1/(2+x^3)dx bounds:0 to .5

2. Originally Posted by cm3pyro
integral of 1/(2+x^3)dx bounds:0 to .5
$\displaystyle \frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$

integrate ...

$\displaystyle \frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$

using the first three terms should get you within 0.00001 of the actual value

3. Originally Posted by skeeter
$\displaystyle \frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$

integrate ...

$\displaystyle \frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$

using the first three terms should get you within 0.00001 of the actual value
i need a maclaurin series

4. Originally Posted by cm3pyro
i need a maclaurin series
what do you think I posted?