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In my book I see:

To integrate the given function we complete the square in the denominator:

$$4x^2 - 4x + 3 = (2x-1)^2 + 2$$

How is it doing this? When I complete the square I get:

$$x^2 - x + \frac{3}{4} = 0$$ $$x^2 - x + \frac{1}{4} = \frac{-1}{2}$$ $$\left(x-\frac{1}{2}\right)^2 = \frac{-1}{2}$$ $$x - \frac{1}{2} = \sqrt{\frac{-1}{2}}$$

now I'm stuck. can someone help to show me how my book gets the $(2x-1)^2 + 2$?

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    $\begingroup$ Why have you set the denominator equal to zero and divided through by 4? This is not a valid procedure. $\endgroup$ – Peter Foreman May 24 '19 at 15:30
  • $\begingroup$ @PeterForeman I thought that's how I complete the square... $\endgroup$ – Jwan622 May 24 '19 at 15:36
  • $\begingroup$ @Jwan622 Completing the square is used in this case to simplify an expression, not to solve an equation. These are different procedures. $\endgroup$ – Peter Foreman May 24 '19 at 15:37
  • $\begingroup$ That's the method used for solving quadratic equations, which is something completely different. $\endgroup$ – KM101 May 24 '19 at 15:37
  • $\begingroup$ This looks like a quadratic no? $\endgroup$ – Jwan622 May 24 '19 at 18:01
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$$4x^2-4x+3=4x^2-4x+1+2=(2x-1)^2+2.$$

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  • $\begingroup$ so this isn't really completing the square right? $\endgroup$ – Jwan622 May 24 '19 at 15:31
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    $\begingroup$ Why not? I wrote your expression as a square plus a number that should be added so that the equality holds. $\endgroup$ – José Carlos Santos May 24 '19 at 15:32
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$4x^2-4x+ 3 = 4x^2-4x + 1 + 2 = (2x-1)^2+2$

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This is exactly the same as the other answers, but I'll try to explain a bit more:

Given expression: $$4x^2 - 4x + 3$$ $$=4\left(x^2-x+\frac{3}{4}\right)$$ $$=4\left((x^2)-2\left(\frac{1}{2}\right)(x)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)$$ $$=4\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+\frac{3}{4}\right)$$ $$=4\left(\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\right)$$ $$=4\left(x-\frac{1}{2}\right)^2+2$$ $$=4\left(\frac{2x-1}{2}\right)^2+2$$ $$=(2x-1)^2+2$$

Because this expression is in the denominator of the integrand, we can further express this as $$(2x-1)^2+(\sqrt2)^2$$ And proceed with the integration, depending upon the expression in the numerator.

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