Bah. Where did the -1 come from?

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http://img48.imageshack.us/img48/6110/qnisstb4.jpg

I always get the positive and negatives confused. I don't have a teacher to guide me so help would be appreciated. The correct answer is in green ink.

Thanks
 
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At one point you have converted 1/(x+2) into (-1)/(2- x). That's wrong. That would be equal to 1/(x-2) not 1/(x+ 2). What you want, rather, is 1/(x+ 2)= 1/(2+x)= 1/(2-(-x)). Then dividing by numerator and denominator by 2: (1/2)/(1-(-x/2)). The "r" in your geometric series is -x/2, not x/2.
 
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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