How do you integrate? #intsqrt(3x^2-3x+1)dx#

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#intsqrt(3x^2-3x+1)dx#

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Answer 1 · 2018-04-13T14:14:42

# 1/4(2x-1)sqrt(3x^2-3x+1)+sqrt3/24ln|(2x-1)/2+sqrt{(3x^2-3x+1)/3}|+C#.

Knowing that, #intsqrt(x^2+a^2)dx=1/2[xsqrt(x^2+a^2)+a^2ln|x+sqrt(x^2+a^2)|+c#.

#:. intsqrt(3x^2-3x+1)dx#,

#=sqrt3intsqrt(x^2-x+1/3)dx#,

#=sqrt3int{(x^2-x+1/4)+1/12}dx#,

#=sqrt3int{(x-1/2)^2+1/12}dx#,

#=sqrt3*1/2[(x-1/2)sqrt(x^2-x+1/3)+1/12ln|(x-1/2)+sqrt(x^2-x+1/3)|]#,

#=1/4(2x-1)sqrt(3x^2-3x+1)+sqrt3/24ln|(2x-1)/2+sqrt{(3x^2-3x+1)/3}|+C#.