evaluate the infinte integral as an infinite series

**Tascja** · March 8th 2009, 04:28 PM

the integral is

3[int.] e^x - 1 / 8x

Im not sure what exactly its asking me... infinite integral using infinite series??

i know that e^x = [Sum.]n=0 to infinite x^n / n!

but i dont know what comes next or even if i have to use that??

An explanation about how to go about solving this would be great

**skeeter** · March 8th 2009, 04:53 PM

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

$e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

$\frac{e^x - 1}{8x} = \frac{1}{8}\left(1 + \frac{x}{2!} + \frac{x^2}{3!} + ... \right)$

now integrate the series on the RHS term for term ...

**Tascja** · March 8th 2009, 05:29 PM

Thank you!

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