##=(x- 1/2) ln(x^2-x+2) – 2x + sqrt 7 arctan ( (2x-1)/sqrt 7)##
##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)##
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