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I= ∫[(x-1)/(x^2-x+1)]/dx
=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
let
x-1/2 = (√3/2)tana
dx = (√3/2)(seca)^2 da
I=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
= (4/3)∫ { [(√3/2)tana- 1/2] /(seca)^2} (√3/2)(seca)^2 da
= 2√3/3∫[(√3/2)tana- 1/2] da
=2√3/3 [-(√3/2)ln|cosa| - a/2] +C
where a= arctan(2√3/3 (x-1/2))
=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
let
x-1/2 = (√3/2)tana
dx = (√3/2)(seca)^2 da
I=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
= (4/3)∫ { [(√3/2)tana- 1/2] /(seca)^2} (√3/2)(seca)^2 da
= 2√3/3∫[(√3/2)tana- 1/2] da
=2√3/3 [-(√3/2)ln|cosa| - a/2] +C
where a= arctan(2√3/3 (x-1/2))
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