int dx qquad ln(x^2-x+2)
we use IBP
int u v' = uv - int u' v
and the trick here is
u = ln(x^2-x+2), u' = (2x-1)/(x^2-x+2)
v' = 1, v = x
so we have
xln(x^2-x+2) - int dx qquad x(2x-1)/(x^2-x+2) qquad star
for int dx qquad x(2x-1)/(x^2-x+2) this becomes:
int dx qquad (2x^2-x)/(x^2-x+2)
next bit is like long division only easier
= int dx qquad (2x^2-2x + 4 + x- 4)/(x^2-x+2)
= int dx qquad 2 + (x- 4)/(x^2-x+2)
now we set it up for a log solution by setting up this pattern: (f'(x)) / f(x)
= 2x + int dx qquad (1/2(2x- 8))/(x^2-x+2)
=2x + int dx qquad (1/2(2x- 1)-7/2)/(x^2-x+2)
= 2x + int dx qquad (1/2(2x- 1))/(x^2-x+2) - 7/2 1/(x^2-x+2)
= 2x + 1/2 ln(x^2-x+2) - 7/2 int dx qquad 1/(x^2-x+2)
if we plug this all into star we have
xln(x^2-x+2) - ( 2x + 1/2 ln(x^2-x+2) - 7/2 int dx qquad 1/(x^2-x+2) )
=(x- 1/2) ln(x^2-x+2) - 2x + 7/2 int dx qquad 1/(x^2-x+2) qquad square
for int dx qquad 1/(x^2-x+2)
we complete the square, looking for a tan sub to finish it off, so
int dx qquad 1/((x-1/2)^2 - 1/4 + 2)
= int dx qquad 1/((x-1/2)^2 + 7/4)
= int dx qquad 1/(7/4 tan^2 phi + 7/4) using the sub (x-1/2)^2 = 7/4 tan^2 phi or (x-1/2) = sqrt 7/2 tan phi
so dx = sqrt 7/2 sec^2 phi \ d phi
=sqrt 7/2 int d phi qquad sec^2 phi \ 1/(7/4 sec^2 phi )
=2/ sqrt 7 int d phi qquad
=2/ sqrt 7 arctan (2 (x-1/2)/sqrt 7)
parking this back into square
=(x- 1/2) ln(x^2-x+2) - 2x + 7/2 2/ sqrt 7 arctan (2 (x-1/2)/sqrt 7)
=(x- 1/2) ln(x^2-x+2) - 2x + sqrt 7 arctan ( (2x-1)/sqrt 7)
