1. LarCalcET6 5.8.037.

Find or evaluate the integral by completing the square:

. . . . .[tex]\displaystyle \int\, \dfrac{1}{\sqrt{\strut -x^2\, -\, 12x\,}}\, dx[/tex]

So when I did complete the square I did

-x^2 -12x

= -(x^2 + 12x)

= -(x^2 +12x +36) - 36

= -36 - (x + 6)^2

I didn't know what to do with the negative so I just ignored it and got the right answer but I feel like I cheated. That negative in front of the x is what F'd me up.

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