How do you integrete this?

#int x/(x^2-x+1)#dx

Is there another method? Not that it was mentioned
Writing x as #1/2# (2x−1)+#1/2#
∫#(2x−12)/(x2−x+1)#+#1/[2(x2−x+1)]#dx

1 Answer
Yahia M.
Apr 3, 2018

You could solve it by trigonometric substitution.
#-ln(cos(tan^-1(2/sqrt(3)(x-1/2))+1/sqrt(3)tan^-1((2/sqrt(3))*(x-1/2))+C#

Explanation:

First, you complete the square in the denominator to have the following formula:

#int(x*dx)/((x-1/2)^2+3/4)#

then by substituting :
#x-1/2=sqrt(3)/2tanz#
#dx=sqrt(3)/2sec^2zdz#
#tan^2z+1=sec^2z#

and by substituting you get the following :

#int((sqrt(3)/2tanz+1/2)*sqrt(3)/2sec^2zdz)/(3/4sec^2z)#
And by simplification you get the following:
#int(tanz+sqrt(3)/3)dz#
=#-ln(cosz)+sqrt(3)/3z+C#
and by substituting back for #z=Tan^-1(2/sqrt(3))(x-1/2)#
You get:

#-ln(cos(tan^-1(2/sqrt(3)(x-1/2))+1/sqrt(3)tan^-1((2/sqrt(3))*(x-1/2))+C#

I hope this was helpful.

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