Calc
posted by Erica .
I have to find the integral of 1/(sq. rt. of (4xx²)) dx. I know I have to complete the square of the denominator, but im confused on how that process works when im only given 4xx². Could you demonstrate how to complete the square for that equation? Thank you so much!

4x  x^2
(x^2  4x)
x^2  4x = 0
x^2  4x + 4 = 0 + 4
(x  2)^2 = 4
(x  2)^2  4
(x  2)^2 + 4
4  (x  2)^2
 = integral sign
 1/(sqrt(4  (x  2)^2) dx
u = ( x  2)
du = dx
 1/(sqrt(4  u^2)) du
= arcsin u/2 + C
= arcsin (x  2)/2 + C
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