Question

Integral of 1/√(3x^2-6x+2)dx

Answer 1

By introducing a substitution,

$t = 3(x-1)^2 - \frac{1}{2}; \\ x = \sqrt{\frac{t+\frac{1}{2}}{3}}+1; \\ dx = \frac{dt}{\sqrt{6}\sqrt{2t+1}}; \\ \smallint{\frac{dx}{\sqrt{3x^2-6x+2}}} = \frac{1}{\sqrt{6*2}}\smallint{\frac{dt}{\sqrt{t+\frac{1}{2}}\sqrt{t-\frac{1}{2}}}} = \frac{1}{\sqrt{12}}\smallint{\frac{dt}{\sqrt{t^2-\frac{1}{4}}}} = \frac{1}{\sqrt{12}}\log{|t+\sqrt{t^2-\frac{1}{4}}|} = \frac{1}{2\sqrt{3}}\log{|3(x-1)^2 - \frac{1}{2}+(x-1)\sqrt{3(3x^2-6x+2)}|}.$