Question #86479
Integral of 1/√(3x^2-6x+2)dx
1
Expert's answer
2019-03-17T16:27:42-0400

By introducing a substitution,

t=3(x1)212;x=t+123+1;dx=dt62t+1;dx3x26x+2=162dtt+12t12=112dtt214=112logt+t214=123log3(x1)212+(x1)3(3x26x+2).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)}|}.


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