Question

Integrate the value and simplify.
∫1/((x-1)√(4x^2-8x+3)) dx

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ (x-1)\sqrt{4x^2-8x+3} }dx\]

Answer 2

Take the integral:
integral 1/((x-1) sqrt(4 x^2-8 x+3)) dx
For the integrand 1/((x-1) sqrt(4 x^2-8 x+3)), complete the square:
= integral 1/((x-1) sqrt((2 x-2)^2-1)) dx
For the integrand 1/((x-1) sqrt((2 x-2)^2-1)), substitute u = 2 x-2 and du = 2 dx:
= 1/2 integral 1/(sqrt(u^2-1) ((u+2)/2-1)) du
For the integrand 1/(sqrt(u^2-1) ((u+2)/2-1)), substitute u = sec(s) and du = tan(s) sec(s) ds. Then sqrt(u^2-1) = sqrt(sec^2(s)-1) = tan(s) and s = sec^(-1)(u):
= 1/2 integral (sec(s))/(1/2 (sec(s)+2)-1) ds
For the integrand (sec(s))/(1/2 (sec(s)+2)-1), do long division:
= 1/2 integral 2 ds
The integral of 2 is 2 s:
= s+constant
Substitute back for s = sec^(-1)(u):
= sec^(-1)(u)+constant
Substitute back for u = 2 x-2:
= sec^(-1)(2 x-2)+constant
Which is equivalent for restricted x values to:
Answer:
-tan^(-1)(1/sqrt(4 x^2-8 x+3))+constant

Answer 3

ok well you beat me to it lol

Answer 4

You two are too kind.