Trouble completing the square on the integral

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Homework Statement



http://c811106.r6.cf2.rackcdn.com/se05h01043.png

Homework Equations





The Attempt at a Solution


The negative sign in front of x^2 is giving me trouble.

When i complete the square i get; \sqrt{}-x^2-4x+4 -4

There's some algebra that i don't understand.
 
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en bloc said:

Homework Statement



http://c811106.r6.cf2.rackcdn.com/se05h01043.png

Homework Equations





The Attempt at a Solution


The negative sign in front of x^2 is giving me trouble.

When i complete the square i get; \sqrt{-x^2-4x+4 -4}

There's some algebra that i don't understand.

First, complete the square for x^2+4x\,. Then take the negative of the result.
 
-x^2- 4x= -(x^2+ 4x)

Now complete the square for x^2+ 4x.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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