Approximating Integrate Using Series Methods

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I need help solving the following (it is due tomorrow :frown: and it just got assigned yesterday ):

use series methods to obtain the approximate value of integral(from 0 to1) of (1-e^x)/x dx

What I have thought of so far is to use the Taylor series approx for e^x and carry that out to a few terms then solve for 1 and 0 and plug it into the original equation but I'm not sure that I'm headed in the right direction. I took a year off from school and now I'm having to relearn a bunch of things that were once simple but now seem quite challenging. Any help is appreciated. Thanks.
 
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Looks good to me. You should get an easy power series to integrate and then you just substitute in 1 and 0.

Mat
 
I plugged in the Taylor series for e^x and took that out to 6 terms then integrated it from 0 to 1 and got 1.262...does that sound about right?
 
The integrand is negative so you should end up with a negative answer.

As an example just using two terms, ex=1+x you would end up with

\int_0^1 \frac{1-(1+x)}{x}dx = \int_0^1 \frac{-x}{x}dx=-1

As you add more terms the integral should become more negative
 
Thanks for pointing that out Office Shredder. There was a negative that I had forgotten to carry over. So I got -1.262 with 6 terms in the e^x Taylor series.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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